The Harmonic Numbers and Series: Torsten Sillke
last update 2000-12-12
H(n) = 1 + 1/2 + 1/3 + ... + 1/n.
H(2n) - H(n) = 1/(n+1) + 1/(n+2) + ... + 1/(2n).
Table:
n : 0 1 2 3 4 5 6 7 8
H(n): 0 1 3/2 11/6 25/12 137/60 49/20 363/140 761/280
Prop A: H(2n) - H(n) = 1 - 1/2 + 1/3 - 1/4 ... + 1/(2n-1) - 1/(2n)
Follows from Induction:
n=0 : H(0) - H(0) = 0
n->n+1: H(2(n+1))-H(n+1) = H(2n) + 1/(2n+1) + 1/(2n+2) - H(n) - 1/(n+1)
= H(2n) - H(n) + 1/(2n+1) - 1/(2n+2)
= 1 - 1/2 + 1/3 - 1/4 ... + 1/(2n+1) - 1/(2n+2).
Prop B: H(2n) - H(n) is monoton increasing and converges to log(2).
The limit of 1 - 1/2 + 1/3 - 1/4 ... was first
calculated by Nicolaus Mercator [Logarithmotechnia, 1667]. He found the series
log(1+x) = x - x^2/2 + x^3/3 - x^4/4 ...
But you can do that without using the Taylor series.
Adding the estimates 1/(n+1) < Integral[1/x, x, {n,n+1}] < 1/n gives
H(2n)-H(n) < Integral[1/x, x, {n,2n}] < H(2n-1)-H(n-1) = H(2n)-H(n)+1/(2n).
Now Integral[1/x, x, {n,2n}] = Integral[1/y, y, {1,2}] via y = n x.
(An interpretation of this transform gives [Yegorov].)
So we get log(2) - 1/(2n) < H(2n) - H(n) < log(2).
As log(t) is defined as the Integral[1/y, y, {1,t}] see [Yegorov].
Prop C: H(n) is integral only for n=1.
The greatest power of 2 not exceeding n don't match any other term (n>1).
Therefore the numerator of H(n) must be odd and the denumerator even.
And further if 2^t <= n then 2^t divides the denumerator.
See [Theisinger], [AAA], [Trigg, Q211], [GKP88 Exc 6.21].
[Nagell] found the generalization that every subsequence of the harmonic
series in arithmetical progression is non integral too except the trivial
case H(1) = 1. For the sum 1 + 1/3 + 1/5 + ... + 1/(2n-1) see [BBB].
Prop D: H(n) diverges.
The text book proof e. g. [Northrop] is the following
H(oo) = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 ...
>= 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 ...
= 1 + 1/2 + 1/2 + 1/2 +
This estimate was first described by Nicole Oresme. [Gericke 1990]
Pietro Mengoli showed this too.
A second proof, a reductio ad absurdum, was given by
[Honsberger 1976, Exercise 10.2] and [Gardner 1981].
Assume that H(oo) is finite then
H(oo) = 1 + 1/2 + 1/3 + 1/4 + ...
= 2/2 + 2/4 + 2/6 + 2/8 + ...
< 1+1/2 + 1/3+1/4 + 1/5+1/6 + 1/7+1/8 + ...
= H(oo)
Other proofs found by Jakob Bernoulli were indirect
or used the geometric series and were more complicated.
[Honsberger 1976, Exercise 10.3] shows with exp(x+y) = exp(x)*exp(y)
that exp(H(n)) > n+1. See also Prop H.
Prop E: Good Bounds
log(n) + 1/2 + 1/(2n) <= H(n) <= log(n) + 1. (for n>=1)
Lower Bound: evaluate Integral[1/x, x, {1,n}] by the trapezoid rule.
log(n) <= 1/2 (1 + 1/n) + (1/2 + 1/3 + ... + 1/(n-1))
As the 1/x is a convex function the sum will be larger. [Conway, Guy]
Upper Bound: Integral[1/x, x, {n-1,n}] > 1/n => log(n) > H(n) - 1
Prop F: Asymptotic
H(n) = log(n) + C_E + 1/(2n) - 1/(12n^2) + 1/(120n^4) - 1/(256n^6) + eps
with 0 <= eps <= 1/(240n^8) and
C_E = 0.5772156649015328606065120900824 the Euler-Mascheroni Constant.
The general form of Euler's summation formula for the function f(x) = 1/x
gives the approximation [Appostol 1999]:
H(n) = log(n) + C_E + 1/(2n) - B_2/(2n^2) - B_4/(4n^4) - B_6/(6n^6) ...
- B_2k/(2k n^(2k)) * eps(n,k) with 0 <= eps(n,k) <= 1
Therefore approximation is as good as the first unused term of the series.
The B_k are the Bernoulli numbers.
Euler himself calculated C_E up to 15 digits.
[Euler, 1732 and 1734], [Burk], [Knuth]. See Steven Finch
http://www.mathsoft.com/asolve/constant/euler/euler.html
Ramanujan offers an asymtotic in m = n(n+1)/2
H(n) = 1/2 log(2m) + C_E + 1/(12m) - 1/(120m^2) + 1/(630m^3) - 1/(1680m^4)
[Ramanujan's Notebooks, Part V, Chap 38, Entry 9, p531]
Prop G: Generating Function
log 1/(1-z) = z + z^2/2 + z^3/3 + z^4/4 + ...
1/(1-z) log 1/(1-z) = H(1) z + H(2) z^2 + H(3) z^3 + H(4) z^4 + ...
Prop H: Combinatorial Interpretation
1 + Sum[ a[n] x^n / n!, n, {1,oo} ] = Exp[ Sum[ c[n] x^n / n!, n, {1,oo} ] ].
a[n] = #{ permutaions of [1..n] } = n!
c[n] = #{ cyclic permutaions of [1..n] } = (n-1)!
These special series give the relation
1/(1-x) = 1 + Sum[ x^n, n, {1,oo} ] = Exp[ Sum[ x^n / n, n, {1,oo} ] ].
Truncating the right sum gives (for x>0) the inequality
1 + Sum[ x^n, n, {1,k} ] < Exp[ Sum[ x^n / n, n, {1,k} ] ].
And for x=1 we get the bound log(k+1) < H(k).
[Stanley]
Stirling numbers (of first kind, unsigned) one gets :
H(n) = S(n+1,2)/n!
Prop I: Inversion
Let Q_n be the least integer m such that H(m) > n.
Find the smallest integer n such that
Q_n <> [exp(n - C_E)+1/2], or prove that no such n exists.
This is an open problem from: Concrete Mathematics, Ex 9.67
[Boas, Wrench 1971][Comtet 1967] showed that the largest integer n with
H(n)<=x is n = [exp(x - C_E)-1/2-3/(2(exp(x-1)-1))] or the same integer
plus one.
H(15092688622113788323693563264538101449859497) is the first value larger 100.
Prop K: digit subsequence
Kempner showed 1914 that the digit 9 free subsequence of
the harmonic series is convergent and bounded by 90.
The set of 9 free numbers has density of order x**log(9/10).
Therefore the sum must converge and we get a not too bad estimate
Integral[1/x x**log(9/10), x, {1,oo}] = 1/log(10/9) = 9.49.
W. Stadje gives bounds for a more general problem.
Prop L: distinct digits
Calculate the sum of the reciprocals of the 8,877,690 positive
integers whose decimal representations contain no repeated digits.
[AMM E2533 by E. S. Pondiczery: Solution AMM 83 (1976) 570]
Prop P: Prime subsequence
Euler showed that the sum 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 +...
over all primes diverges. Elementary proofs are given be
[Erd"os], [Clarkson], and [Niven].
If we know the density of the primes is 1/log(n) then
we can approximate the above sum upto 1/n as
Integral[1/x 1/log(x), x, {1,n}] = log(log(n)).
For a proof see [Apostol, Chap 4.8].
The sum of 1/p over all primes p <= n is
log(log(n)) + 0.2614972128 + O(1/log(n)) where the constant
is called Meissel Mertens constant. See Steven Finch
http://www.mathsoft.com/asolve/constant/hdmrd/hdmrd.html
The proof of [Niven] gives a good bound for the partial series.
Let Hp(n) = Sum_{p <= n with p prime} 1/p
and Hs(n) = Sum {s <= n with s squarefree} 1/s then using exp(x)>1+x for x>0
exp(Hp(n)) = Prod exp(1/p) > Prod (1+1/p) >= Hs(n)
As Hs(n) * zeta(2) >= H(n) and zeta(2) = 1 + 1/4 + 1/9 + 1/16 + ... < 2
we get Hp(n) >= log(H(n)/2) >= log(log(n) - log(2).
Prop Q: arithmetic progressions
Is it true that whenever {a(n)} is a sequence of positive integers
such that Sum[ 1/a(n), n, {1,oo}] diverges, then the sequence {a(n)}
contains arbitrarily long arithmetic progressions.
[a problem of Paul Erd\Hos worth $3000]
Prop R: generalize H(n) to real numbers
H(x) := (1/1 - 1/(1+x)) + (1/2 - 1/(2+x)) + (1/3 - 1/(3+x)) + ...
Psi(x) = Gamma'(x)/Gamma(x)
H(n) - C_E = Gamma'(n+1)/Gamma(n+1)
H(x) = Integral[(1-(1-t)^x)/t, t, {0, 1}]
Prop W: p^2 divides the numerator of H(p-1) for each prime p>3.
This is Wolstenholme's theorem.
Prop Z: Zeta function
zeta(x) = 1/1^x + 1/2^x + 1/3^x + 1/4^s + ...
The zeta function (and its analytic extension) is defined
on the entire complex plane. At x=0 it has a pol of first order.
Euler in 1749 found the infinite product representation
1/zeta(x) = (1-1/2^x)(1-1/3^x)(1-1/5^x)(1-1/7^x)(1-1/11^x)...
where the product goes about all prime numbers. This is the
algebraic form to say that the prime factorization is unique.
In (1737) he remarked that he can use this relation to show (Prop P).
This shows that there are much more primes than square numbers as
zeta(2) = 1 + 1/4 + 1/9 + 1/16 + ... = pi*pi/6 which he found in (1734).
see [Ayoub]
Applications:
- infinite offset [Trigg, Q 52 Piled Dominoes]
A nice picture of this can be found in Robert M. Dickau page
"Harmonic numbers and the book-stacking problem"
http://forum.swarthmore.edu/advanced/robertd/harmonic.html
- quicksort (expected number of comparisons)
- deep in the dessert [Fine 1947]
- coupon collection problem (occupation problem)
- expected number of cycles in a random permutation
- expected number of left-right-minima (minimim algorithm)
- number of permutations with 2 orbits: |stirling_first(n+1,2)| = n! H(n)
- The product (1 - 1/p) over all primes is 1/H(oo). [Euler]
Therefore the number of primes is infinite.
(See [Ribenboim] or a detailed explanation in [Rademacher, Toeplitz])
- Show C_E < T(n,m) = H(n) + H(m) - H(nm) <= 1 for n,m >= 1. See [CCC].
- Special determinant: Putnam competition 1992, Problem B5
Let D_n denote the value of the (n-1) by (n-1) determinant
| 3 1 1 1 ... 1 |
| 1 4 1 1 ... 1 | Is D_n/n! bounded?
| 1 1 5 1 ... 1 |
| 1 1 1 6 ... 1 |
| . . . . ... . |
| 1 1 1 1 ... n+1 |
Solution: No, as D_n/n! = H(n).
| b+a a a a |
| a c+a a a |
| a a d+a a | =
| a a a e+a |
| b+a a a a a |
| a c+a a a a |
| a a d+a a a | =
| a a a e+a a |
| 0 0 0 0 1 | subtract the last column from every other
| b 0 0 0 a |
| 0 c 0 0 a |
| 0 0 d 0 a | =
| 0 0 0 e a |
| -1 -1 -1 -1 1 | eleminate the -1 of the last row by adding rows
| b 0 0 0 a |
| 0 c 0 0 a |
| 0 0 d 0 a | =
| 0 0 0 e a |
| 0 0 0 0 1+a/b+a/c+a/d+a/e | product of the diagonal elements
bcde(1+a/b+a/c+a/d+a/e) =
sigma_4(a,b,c,d,e) = abcde(1/a + 1/b + 1/c + 1/d + 1/e)
- a series [Gould 1961]
There is an analytic way to approach summation problems:
If f(x) = Sum_{k>=0} a[k] x^k, and this series converges for x=x0,
then show that
Sum_{k>=0} a[k] x0^k H(k) = Integral_{0 to 1} (f(x0) - f(x0 y))/(1-y) dy
- IMO 5 competition
Let m and n be positive integers such that:
m/n = 1 - 1/2 + 1/3 - 1/4 + ... - 1/1318 + 1/1319.
Prove that m is divisible by 1979.
Solution:
Let H(n) = 1/1+1/2+1/3+...+1/n; let A(n) = 1/1-1/2+1/3-1/4+...+-1/n.
A cool trick gives us A(2n+1)=H(2n+1)-H(n); if you haven't seen it
before, it goes like this:
1/2+1/4+1/6+...+1/2n = H(n)/2, by the distributive property.
1/1+1/3+1/5+...+1/(2n+1) = H(2n+1) - H(n)/2, since we're subtracting
the even denominators away from H(2n+1).
Subtracting away the even denominators again from the equation in the
previous sentence, we get 1/1-1/2+1/3-1/4+1/5-1/6+...-1/2n+1/(2n+1) =
H(2n+1) - H(n)/2 - H(n)/2. Simplifying both sides, we get A(2n+1) =
H(2n+1) - H(n).
This means that m/n =
= A(1319)
= H(1319)-H(659)
= 1/660+1/661+1/662+...+1/1319
= Sum_{i=660}^{1319} [1/i]
= Sum_{i=0}^{659} [1/(660+i)]
= Sum_{i=0}^{329} [1/(660+i) + 1/(660+659-i)]
= Sum_{i=0}^{329} [(1319-i+660+i)/(660+i)(1319-i)]
= Sum_{i=0}^{329} [(1979)/(660+i)(1319-i)]
= 1979 * Sum_{i=0}^{329} [1/(660+i)(1319-i)]
= 1979 * (r/s)
...where r and s are derived the stupid way: multiply each summand's
numerator and denominator by the denominators of the other 329
summands. That is, s = Prod_{i=0}^{329} [(660+i)(1319-i)] and r is
some uninteresting integer.
We have 1979*r*n=s*m, meaning 1979|sm, or 1979|660*661*662*...*1319*m.
Since 1979 is prime, it must divide at least one of the multiplicands
on the right-hand-side, but most of them are less than 1979; the only
option we have is 1979|m. QED.
--brian
References:
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Proofs from THE BOOK,
Springer, New York, Berlin, 1998.
- Tom M. Apostol;
Introduction to Analytic Number Theory,
Springer Verlag, New York, Berlin, 1980
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Another elementary proof of Euler's formula for \zeta(2n),
American Mathematical Monthly 80 (1973) 425-431
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A proof that Euler missed: Evaluating zeta(2) the easy way,
Mathematical Intelligenzer 5 (1983) 59-60
see [Aigner, Ziegler; p29-31]
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An Elementary View of Euler's Summation Formula,
American Mathematical Monthly 106 (1999) 409-418
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Euler and the zeta function,
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Thinning out the Harmonic Series,
Mathematics Magazine 68:4 (Oct. 1995) ???
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Sums of generalized harmonic series and volumes,
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partial sums of the harmonic series,
American Mathematical Monthly 78 (1971) 864-870
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Elementary evaluation of \zeta(2n),
Mathematics Magazine 48 (1975) 148-154
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Euler's constant,
The College Mathematics Journal, 16 (1985) 279
- the Euler-Mascheroni Constant
- H. L. L. Busard;
Nicole Oresme: Questiones super geometriam Euclides,
Leiden 1961
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American Mathematical Monthly 20 (1913)
5-14, 35-47, 75-84, 107-117, 148-151, 173-182, 205-210
(history of the concepts of logarithms and exponetials)
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On the series of prime reciprocals,
Proceedings of the American Mathematical Society 17 (1966) 541
MR 32 # 5573
see [Apostol 1980; chap 1.6]
(1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + ... diverges)
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Summing the Harmonic Series,
Mathematics Teacher, 77:3 (Mar. 1984) 195-196
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About \Sum 1/n,
American Mathematical Monthly 74 (1967) 209
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The Book on Numbers,
Copernicus an imprint of Springer Verlag, 1996
(german: Zahlenzauber, Birkh\"auser, 1997)
- Chapter 9: some transcendent numbers
Section: the harmonic numbers
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On the divergence of the harmonic series,
by Don Crossfield. - Washington, D. C. : State Univ., 1991. - [2 Bl.].
(Mathematics notes from Washington State University ; 132 = Vol. 34,2)
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Summa summarum,
Teubner Verlagsgesellschaft, Leipzig, 1986
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- vom Neunerfresser
- der schiefe Turm
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Series expansions from Mengoli to Newton (Italian),
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The Bernoullis and the harmonic series.
The College Mathematical Journal 18:1 (1987) 18-23
MR 88e:01017 01A45
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Mathematical Problems: An Anthology
(Revised english edition by Richard A. Silverman)
Gordon and Breach, New York, 1967
- prob 25: (Prop A)
- prob 80: (Prop C)
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\"Uber die Reihe Sum 1/p,
Mathematica, Zutphen B 7 (1938) 1-2.
see [Aigner, Ziegler; p6]
(1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + ... diverges)
- Leonhard Euler;
Methodus generalis summandi progressiones,
Commentarii academiae scientiarum Petropolitanae 6 (1732) 68-97.
reprinted in: Opera Omnia, series 1, volume 14, 42-72
- Leonhard Euler;
De progressionibus harmonicis observationes,
Commentarii academiae scientiarum Petropolitanae 7 (1734) 150-161.
reprinted in: Opera Omnia, series 1, volume 14, 87-100
- Leonhard Euler;
Methodus universalis serierum convergentium summa quam proxime inveniendi,
Commentarii academiae scientiarum Petropolitanae 8 (1736) 3-9.
reprinted in: Opera Omnia, series 1, volume 14, 101-107
(Euler Maclaurin summation formula)
- Leonhard Euler;
Methodus universalis series summandi ulterius promota,
Commentarii academiae scientiarum Petropolitanae 8 (1736) 147-158.
reprinted in: Opera Omnia, series 1, volume 14, 124-137
(Euler Maclaurin summation formula)
- Charles Vanden Eynden;
Proofs that $\Sigma \frac (1)(p)$ diverges,
American Mathematical Monthly 87 (1980) 394-397
- N. J. Fine;
The jeep problem,
American Mathematical Monthly 54 (1947) 24-31
(reprinted in: D. Gale, Tracking the Automatic Ant, 1998, Appendix 2)
- Martin Gardner;
M. Gardner's Sixth Book of Mathematical Games from Scientific American,
Freeman, 1971, San Francisco
- Chapter 17: Limits of Infinite Series
- Martin Gardner;
Puzzles from other worlds,
Woods End, 1981
(german: Denkspiele von anderen Planeten, Hugendubel, 1986)
- Problem 28: Dancing Superballs
(The harmonic series is divergent, reductio ad absurdum)
- Helmuth Gericke;
Mathematik in Antike und Orient, 1984
Mathematik im Abendland, 1990
(reprinted in one volume: Fourier Verlag, Wiesbaden, 1994, 3. Aufl.)
- H. W. Gould;
Some relations involving the finite harmonic series,
Mathematics Magazine 34 (1961) 317-321
- [GKP88] Ron L. Graham, Donald E. Knuth, Oren Pataschnik;
Concrete Mathematics,
Addison Wessley, Reading, 1994, 2nd Ed.
- Chapter 6.3: Harmonic Numbers
- Ross Honsberger;
Ingenuity in Mathematics,
New Mathematical Library, Vol 23,
Math. Assoc. America, Washington, DC,
- (1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + ... diverges)
- Ross Honsberger;
Mathematical gems, II
The Dolciani Mathematical Exposition 2,
Math. Assoc. America, Washington, DC, 1976;
(german translation: Mathematische Juwelen, Vieweg, 1982)
MR 57 #11978
- chap 10: an interesting series, pp. 98--103,
- Alan Hausrath, Bradley Jackson, John Mitchem, Edward Schmeichel;
Gale's round-trip jeep problem,
American Mathematical Monthly 102:2 (Apr 1995) 299-309
- Frank Irwin;
A Curious Convergent Series,
American Mathematical Monthly 23 (1916) 149-152
(22.4 < the 9 free subsequence < 23.3. improves [Kempner])
- A. J. Kempner;
A Curious Convergent Series,
American Mathematical Monthly 21 (1914) 48-50
(the 9 free subsequence is bounded by 90)
- Frank K. Kenter;
A Matrix Representation for Euler's Constant, \gamma,
American Mathematical Monthly 106 (1999) 502-504
- Donald E. Knuth;
Euler's constant to 1271 places,
Mathematics of Computation 16 (1962) 275-281
- Donald E. Knuth;
The Art of Computer Programming, Vol 1,
Fundamental Algorithms,
Addison Wesley 1968 (2nd edition 1973)
- chapter 1.2.7: Harmonic numbers
- excercise 1.3.2.16-17 fixed precision harmonic series
- W. G. Leavitt;
The sum of the reciprocals of the primes,
Two Year College Mathematics Journal 10, 198-199
- Colin Maclaurin;
A Treatise of Fluxions,
Edinburgh, 1742
(Euler Maclaurin summation formula)
- T. Nagell;
>> $\sum\sb {k=0}\sp n(m+kd)\sp {-1}$ cannot be an integer <<
Skr. Norske Vid. Akad. Kristiania. I. 1923, no. 13, 10--15 (1924)
- Ivan Niven;
A Proof of the Divergence of Sum 1/p,
American Mathematical Monthly 78 (1971) 272-273
(1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + ... diverges)
- Eugene P. Northrop;
Riddles in Mathematics, a Book of Paradoxes,
D. Van Nosttrand Company, New York, 1944
- Chapter 7: Outward Bounds (Paradoxes of the Infinite)
- R. Reiff;
Geschichte der unendlichen Reihen, M\"unchen, 1889,
reprinted: Wiesbaden, S\"andig, 1969
- Hans Rademacher, Otto Toeplitz;
Von Zahlen und Figuren,
Heidelberger Taschenb\"ucher Band 50,
Springer Verlag, Berlin, New York, 1968 (1st edition 1930)
- chapter 17b: Eulers Beweis f\"ur das Nichtabbrechen der Primzahlreihe
- Paulo Ribenboim;
The New Book of Prime Number Records.
Springer, New York, 1996, 3rd ed.
ISBN 0-387-94457-5
- Neville Robbins;
Revisiting an Old Favorite: \zeta(2m),
Mathematics Magazine, 72:4 (Oct. 1999) 317-319
- G. R. Sanchis;
Swapping Hats: A Generalization of Montmort's Problem
Mathematics Magazine 71:1 (1998) 53-57
(Let P_m(N) be the probability that m is the minimal cycle length
of the cycles of a random permutation of 1..N.
Thm: The limit (N->oo) of P_m(N) is exp(- H(m)) with
The proof is elementary by inclusion exclusion.)
- W. Stadje;
Konvergenz von Teilen der harmonischen Reihe. (German)
[Convergence of parts of the harmonic series],
Elemente der Mathematik 46:2 (1991), 51-54
- Richard P. Stanley;
Exponential Structures,
Studies in Appl. Math. 59 (1978) 73-82
#MR 58.262 (E. A. Bender)
- Hugo Steinhaus;
100 neue Aufgaben,
Urania Verlag, 1973, Leipzig-Berlin
- problem 92: Eine M\"unze jenseits der Tischkante (overhanging dominoes)
- L. Theisinger;
Bemerkungen \"uber die harmonische Reihe,
Monatshefte f\"ur Mathematik und Physik 26 (1915) 132-134
- The denonimator of H_n is a multiple of 2^floor(log2(n)).
- Charles W. Trigg;
Problem 52: Overhanging dominoes, (proposer: R. T. Sharp)
Pi Mu Epsilon Journal 1:10 (April 1954) 411-412
- Charles W. Trigg;
Mathematical Quickies, 1967
- John Wallis;
Arithmetica infinitorum, 1655/56
(show how to integrate x**m)
- Andrey Yegorov;
Squaring the hyperbola,
A different approach to logarithms and exponents,
Quantum (March/April 1997) 27-31, 56
- AAA
(Prop C)
American Mathematical Monthly 41 (Jan. 1934) 48
- BBB
American Mathematical Monthly 67 (1960) 924-925
- CCC
American Mathematical Monthly 70 (1963) 575-577
Links:
- Kevin Brown;
Bernoulli Numbers and Harmonic Series,
http://www.seanet.com/~ksbrown/kmath284.htm
- calculates Euler's constant with the Euler MacLaurin summation formular
- calculates Euler's constant via H(2^n) as a linear recursion approximation
where we only have to solve a linear system of equations
- Kevin Brown;
Legendre's Prime Number Conjecture,
http://www.seanet.com/~ksbrown/kmath032.htm
- lim_{x->oo} 1/log(x) * Prod_{p <= x and p prime} (1 + 1/p) =
6 exp(C_E) / pi^2 = 1.082762...
- FunFacts;
Thinned-Out Harmonic Series,
http://www.math.hmc.edu/funfacts/ffiles/20005.3.shtml
- subset of numbers without the digit 9.
- John H. Webb;
In Perfect Harmony,
http://pass.maths.org/issue12/features/harmonic/
- many applications of the logarithmic groth of the harmonic series.
- records, destruction tests, shuffling cards, crossing the desert
Timeline:
- Leonardo von Pisa (1170/80-1240)
- Nicole Oresme (1320/25-1382)
- John Napier (1550-1617)
- Gregorius a St. Vincentio (1584-1667)
- John Wallis (1616-1703)
- Nicolaus Mercator (1620-1687)
- Pietro Mengoli (1626-1686)
- Gottfried Wilhelm Leibniz (1646-1716)
- Johannes (I) Bernoulli (1667-1748)
- Leonhard Euler (1707-1783)
http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Mengoli.html
http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Mercator_Nicolaus.html
http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Oresme.html
http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Bernoulli_Johann.html
Appendix:
(A) H(n) diverges. The indirect proof of Jakob Bernoulli 1689 [Dunham].
Start with relation
1/a = 1/(a(a+1)) + 1/((a+1)(a+2)) + 1/((a+2)(a+3)) + ...
= ( 1/a - 1/(a+1) ) + ( 1/(a+1) - 1/(a+2) ) + ( 1/(a+2) - 1/(a+3) ) + ...
Then determine the sum of the following infinite table of positive numbers
by summing the columns first and then by summing the rows first.
1/2 1/3 1/4 1/5 ...
+------------------------------------------
1 | 1/(1*2) 1/(2*3) 1/(3*4) 1/(4*5) ...
1/2 | 1/(2*3) 1/(3*4) 1/(4*5) ...
1/3 | 1/(3*4) 1/(4*5) ...
1/4 | 1/(4*5) ...
... | ...
Therefore 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... = 1/2 + 1/3 + 1/4 + 1/5 + ...
But this can't hold if the sum is finite.
(B) The series of inverse squares is convergent.
Sum 1/n^2 < 1 + Sum 1/(n*(n+1)) = 2
(C) The inverse squarefree number series
xi(s) = (1+1/2^s)(1+1/3^s)(1+1/5^s)(1+1/7^s)(1+1/11^s)(1+1/13^s)(1+1/17^s)...
= 1^s + 1/2^s + 1/3^s + 1/5^s + 1/6^s + 1/7^s + 1/10^s + 1/11^s + 1/13^s + ...
(The product is over all primes and the sum over all squarefree numbers.)
As 1/zeta(s) = (1-1/2^s)(1-1/3^s)(1-1/5^s)(1-1/7^s)(1-1/11^s)(1-1/13^s)...
we get the relation xi(s)/zeta(s) = 1/zeta(2s) by binomial relation.
The relation zeta(s) = xi(s) zeta(2s) is the analytic way of saying that
each positive integer is the product of a square and a squarefree number
in a unique way.
Reviews:
--------
99c:26019 26D15 (05A20 11B68)
Stankovi\'c, Miomir S.(YU-NISO); Dankovi\'c, Bratislav M.(YU-NISEE-A);
Tri\v ckovi\'c, Slobodan B.(YU-NISCE)
Some inequalities involving harmonic numbers. (English. English summary)
Recent progress in inequalities (Ni\v s, 1996), 493--498, Math. Appl., 430,
Kluwer Acad. Publ., Dordrecht, 1998.
Summary: "In this paper we consider some inequalities for convex functions
and derive sharper lower and upper bounds for harmonic numbers. Using the
Hadamard integral inequality, we get some better estimates.
Also, we give a few applications to some functions."
91c:05021 05A19 (11B37 68R05)
Spiess, Jrgen(D-BRNS)
Some identities involving harmonic numbers.
Math. Comp. 55 (1990), no. 192, 839--863.
This paper is a systematic investigation of the properties of the
generalized harmonic numbers $H\sp {(r)}\sb n=\sum\sp n\sb {k=1}k\sp {-r}$.
The monomials of the sums are products of binomial coefficients and
generalized harmonic numbers. The results obtained are reduction formulas
for these sums. The main interest of this paper is that it gives formulas
which can be useful in computer algebra systems. In fact the proofs are
constructive and thus allow one to compute effectively the polynomials
on which the reduction depends.
Reviewed by M.-P. Delest
90i:40005 40A25 (26A06 33A15)
Savio, Dominic Y.(1-RI); Lamagna, Edmund A.(1-RI); Liu, Shing-Min(1-RI)
Summation of harmonic numbers. Computers and mathematics (Cambridge, MA, 1989),
12--20, Springer, New York-Berlin, 1989.
The authors give formulas for a summation of expressions of the form
$p(i)H\sb i$ and $f(i)H\sb i$ where $p$ is a polynomial, $f$ is a
rational function and $H\sb i=1+(\tfrac12)+\cdots+(\tfrac1i)$ is a
harmonic number. The formulas are obtained by employing a relationship
between harmonic numbers and polygamma functions (i.e., logarithmic
derivatives of the gamma function) and by using R. Moenck's algorithm
[cf. in Proceedings of the MACSYMA User's Conference (Berkeley, CA, 1977),
225--236; per bibl.].
\{For the entire collection see MR 90d:00036\}.
Reviewed by Maciej Sablik
87e:05014 05A19 (05A30 33A30)
Andrews, George E.(1-PAS); Uchimura, K.
Identities in combinatorics. IV. Differentiation and harmonic numbers.
Utilitas Math. 28 (1985), 265--269.
The authors give some examples of series whose $k$th term contains
$H\sb k=\sum\sp k\sb {j=1}j\sp {-1}$ or a $q$-analogue
$H\sb k(q)=\sum\sp k\sb {j=1}q\sp j(1-q\sp j)\sp {-1}$, and
show how to obtain series transformations or summations of these series
by use of differentiation. Their method is the one that should always be
the starting place for trying to understand many series. Write the series
as a hypergeometric or basic hypergeometric series to see what it is, and
then consider the few things that can be done to transform such series.
Readers who have not learned this trick should see Section 5 of Andrews'
paper [SIAM Rev. 16 (1974), 441--484; MR 50 #5044].
\{Part III has been reviewed [MR 86i:05020].\}
Reviewed by R. A. Askey
56 #143 05A10
Zave, Derek A.
A series expansion involving the harmonic numbers.
Information Processing Lett. 5 (1976), no. 3, 75--77.
The author gives an expansion of $(1-z)\sp {-(m+1)}[\ln(1-z)\sp {-1}]\sp n$
in terms of binomial coefficients and the cycle index polynomial of the
symmetric group (with alternating character) evaluated at differences of
generalized harmonic numbers. Its use in asymptotic analysis is demonstrated
by a computation of the covariance of the number of left-to-right maxima and
the number of right-to-left maxima in a random permutation.
Reviewed by Dennis White
CMP 1 410 941 (97:01) 05A19
Xu, Lizhi(PRC-DUT-IM)
A kind of combinatorial sums involving the harmonic numbers. (English. English,
Chinese summary) J. Math. Study 28 (1995), no. 1, 11--13.
CMP 1 144 450 (92:06) 40A99
Starc, Zdravko; Arslanagi\'c, \v Sefket
Harmonic numbers. (Serbo-Croatian)
Matematika (Zagreb) 18 (1989), no. 3, 52--61.
99f:26039 26D15
Simi\'c, Slavko
Best possible bounds and monotonicity of segments of harmonic series. II.
(English. English summary)
Mat. Vesnik 50 (1998), no. 1-2, 5--10.
Let $s\sb n(p,q,A,B) = {1 \over np+A} + {1 \over np+A+1} + \cdots + {1 \over nq+B}$, where
$q>p\geq 1$, $B\geq A-1 \geq 0$. The author gives an answer to a conjecture formulated in his
earlier paper [Part I, Mat. Vesnik 38 (1986), no. 3, 331--336; MR 88d:40006] concerning
monotonicity of this sequence, thus obtaining exact bounds on $s\sb n(p,q,A,B)$.
Reviewed by Kirill A. Kopotun
88d:40006 40A05
Simi\'c, S.
Monotonicity of segments of harmonic series. I. (English. Serbo-Croatian summary)
Mat. Vesnik 38 (1986), no. 3, 331--336.
Let $$S(a\sb n,b\sb n)=\frac{1}{a\sb n}+\frac{1}{a\sb n+1}+\frac{1}{a\sb n+2}+\cdots+
\frac{1}{b\sb n},$$ where $(a\sb n)$ and $(b\sb n)$ are monotonic increasing sequences of
integers and $a\sb n\le b\sb n$. As the question of the best possible bounds is in close connection
with the monotonicity of the sequences, the author establishes three propositions pertaining to
monotonicity of the sequences $S(a\sb n,b\sb n)$. The results proved are sufficient for solving the
question of best possible bounds for a wide class of sequences $S(a\sb n,b\sb n)$.
Reviewed by Shri Nivas Bhatt
98c:26016 26D15
Yang, Bi Cheng; Wang, Gen Qiang
Some inequalities for harmonic series. (Chinese. English, Chinese summary)
J. Math. Study 29 (1996), no. 3, 90--97.
Summary: "In this paper, improving on the Euler-Maclaurin summation formula,
we construct some inequalities for harmonic series and refine some classical
inequalities."
Reviewed by Yong Ping Liu
95m:40006 40A25 (65B10)
Walls, Gary L.(1-SMS)
Partial sums of the harmonic series.
Math. Comput. Ed. 27 (1993), no. 1, 33--36.
From the text: "This note employs techniques used in proving the integral test to help compute
partial sums of the harmonic series to a higher degree of accuracy than one would expect possible
with only a hand-held calculator. The technique could be used on any series for which the integral
test can be applied. In the case of a convergent series this could enable someone to determine the
sum of an infinite series to a higher degree of accuracy than would be possible by simply adding a
large number of terms. In the case of a divergent series it is interesting to be able to compute
accurately the sum of many more terms than the fastest computer could compute."
94k:11012 11B05
Powell, Barry J.; \v Salßt, Tibor(SK-KMSK-AL)
Convergence of subseries of the harmonic series and asymptotic densities of sets of positive
integers. (English. English summary)
Publ. Inst. Math. (Beograd) (N.S.) 50(64) (1991), 60--70.
Summary: "We investigate the relation between the convergence of subseries $\sum\sp \infty\sb
{n=1} m\sp {-1}\sb n$ of the harmonic series $\sum\sp \infty\sb {n=1} n\sp {-1}$ and the
asymptotic densities $d(M)$ of sets $M=\{m\sb 1 0$. In
Theorem 8 we give a new proof of the known result that $\sum\sb {m\in M}m\sp {-1}<+\infty$ if
and only if $\sum\sp \infty\sb {n=1} M(n)/n\sp 2<+\infty$. We thus give new formulations of
well-known principles of analytic number theory.
"Numerous remarks and examples are provided throughout the paper in supplement to and
clarification of the main theorems."
94j:11022 11B68 (11M06)
Beukers, Frits(NL-UTRE); Kolk, Johan A. C.(NL-UTRE); Calabi, Eugenio(1-PA)
Sums of generalized harmonic series and volumes.
Nieuw Arch. Wisk. (4) 11 (1993), no. 3, 217--224.
It is a classical result that for natural numbers $n$ one has $$\sum\sp \infty\sb {k=0}(-1)\sp
{nk}(2k+1)\sp {-n}=\delta\sb n(\pi/2)\sp n,$$ where $\delta\sb n\in\bold Q$. Indeed, $\delta\sb n$
is related to the Bernoulli number $B\sb n$ or to the Euler number $E\sb {(n-1)/2}$ according as
$n$ is even or odd. The aim of the authors is to give a proof of this result that "requires no special
knowledge beyond what is included in a good second-year calculus class". The number $\delta\sb
n$ is interpreted as the volume of a certain $n$-dimensional convex polytope with rational
vertices, and its value can be computed from the Taylor expansion $$\tfrac12({\rm sec}\,t+{\rm
tan}\,t)=\sum\sp \infty\sb {n=1}\delta\sb n t\sp {n-1}.$$
Reviewed by Matti Jutila
92j:41042 41A60 (40A25)
DeTemple, Duane W.(1-WAS-AM); Wang, Shun-Hwa(1-WAS-AM)
Half integer approximations for the partial sums of the harmonic series.
J. Math. Anal. Appl. 160 (1991), no. 1, 149--156.
The authors obtain an asymptotic estimate for $S\sb n=\sum\sp n\sb {j=1}(1/j)$ in terms of
$n+\tfrac12$. Bounds for the Comtet function [see R. P. Boas, Math. Comp. 31 (1977), no. 137,
257--264; MR 55 #13730] and computation of its asymptotic expansion are also given. This
extends some earlier results [see Boas, op. cit.; Boas and J. W. Wrench, Jr., Amer. Math. Monthly
78 (1971), 864--870; MR 44 #7179; S. M. Zemyan, Proc. Amer. Math. Soc. 95 (1985), no. 1,
83--86; MR 86m:40002].
Reviewed by H. P. Dikshit
55 #13730 65B15
Boas, R. P., Jr.
Growth of partial sums of divergent series.
Math. Comp. 31 (1977), no. 137, 257--264.
Let $f$ be a positive decreasing function, $f(\infty)=0$, $\vert f\sp {(n)}(x)\vert (n=1,2,3)$
decreasing for large $x$ and of $O(f(x)x\sp {-n})$, $\sum f(n)$ divergent, $\varphi(x)=\int\sb 1\sp
xf(t)\,dt$, $\psi(y)$ the inverse of $y=\varphi(x)$, $\psi"'$ eventually monotone, $s\sb n=\sum\sb
1\sp nf(k)$, $\gamma=\lim\sb {n\rightarrow\infty}(s\sb n-\varphi(n))$, $A=2,3,\cdots$; $n\sb A$
the smallest integer $n$ such that $s\sb n\geq A$ but $s\sb {n-1}{\textstyle\frac
1{2}}+n\sp {-1}$. There are also numerical tables giving $e\sp {A-\gamma}$ to 16 significant
figures, $A=1$ through 20 and $A=100$, and values of the partial sums $S\sb n$ of $\sum 1/k$ to
10 decimal places for various values of $n$.
Reviewed by F. P. Cass
86m:40002 40A05
Zemyan, Stephen M.(1-PASMA)
On two conjectures concerning the partial sums of the harmonic series.
Proc. Amer. Math. Soc. 95 (1985), no. 1, 83--86.
It is known that $S\sb n=\sum\sp n\sb {m=1}1/m$ is never an integer for $n>1$.
As $S\sb n$ diverges, for a positive integer $k$ there exists a unique integer
$n\sb k$ such that $S\sb {n\sb k-1}1, -2/e\sp {3(k-\gamma)}-\frac 12<
n\sb k-(e\sp {-k-\gamma}- 1/(24e\sp {k-\gamma}))<\frac 12+2/e\sp {3(k-\gamma)}$.
He further proposes a conjecture concerning the amount by which
$S\sb {n\sb k-1}$ and $S\sb {n\sb k}$ differ from $k$.
Reviewed by Shri Nivas Bhatt
92h:11009 11A63
Stadje, W.(D-OSNB)
Konvergenz von Teilen der harmonischen Reihe.
(German) [Convergence of parts of the harmonic series]
Elem. Math. 46 (1991), no. 2, 51--54.
As is well known, the harmonic series, $\sum\sp {+\infty}\sb {n=1}1/n$,
is divergent. However, if we let $N$ denote the set of integers whose
decimal representation has no $9$'s in its digits, then A. J. Kempner
has shown that $\sum\sb {n\in N}1/n$ is convergent [Kempner, Amer. Math.
Monthly 21 (1914), no. 1, 48--50]. In R. Honsberger's book Mathematical
gems, II [see pp. 98--103, Math. Assoc. America, Washington, DC, 1976;
MR 57 #11978] a simple proof of this fact is given as well as the estimate
$\sum\sb {n\in N}1/n<90$. Honsberger's book also contains references to
related results for other missing digits. In this paper the author wishes
to generalize these results.
Let $d\geq 2$ be a natural number, let $k$ be a nonnegative integer and let
$j$ be in the set $\{0,1,\cdots,d-1\}$. Let $M$ be the set of all natural
numbers whose $d$-adic expansions contain the digit $j$ at most $k$ times.
Then the following result is proved. Theorem: We have
$$\sum\sb {n\in M}\frac 1n < d\sp {k+1}(1+\log(d\sp {k+1}-1)) <
d\sp {k+1}(1+(k+1)\log d).$$
The proof is a matter of checking the sums over the stretches without
the offending digit and estimating the remains.
As an example, Kempner's result is the case $d=10$, $j=9$ and $k=0$.
The theorem gives $\sum\sb {n\in N}1/n < 10(1+\log 9)<32$.
Reviewed by Don Redmond
89m:05010 05A19
Atanassov, Krassimir T.
Notes on the harmonic series.
Bull. Number Theory Related Topics 10 (1986), no. 2, 10--20.
From the text: "Let $H\sb k=\sum\sb {i=1}\sp k1/i$.
There exists for each integer $k>1$ a corresponding $n\sb k$ such that
$H\sb {n\sb k-1}0$ by the integral
$\Gamma(x)=\int\sb 0\sp \infty e\sp {-t} t\sp {x-1}dt$. Consider
$\psi(x)=\Gamma'(x)/\Gamma(x)$, $x>0$. The fundamental relations
$\Gamma(x+1)=x\Gamma(x)$ and $-\Gamma'(1)=C$ (the Euler constant)
imply at once the equality (1) $H\sb n=1+\tfrac 12+ \cdots+1/n=\psi(n+1)+C$,
$n=1,2,\cdots$.
In this paper, the author generalises $H\sb n$ by defining
$H\: (0,\infty)\to\bold R$, $H(x)=\psi(x+1)+C$, $x\in(0,\infty)$.
Further he obtains upper and lower estimates for $H(x)$,
and obtains a new proof of an interesting theorem by Atanassov.
Reviewed by V. L. Deshpande
88m:40004 40A99
Atanassov, K. T.
Remark on the harmonic series.
C. R. Acad. Bulgare Sci. 40 (1987), no. 5, 25--28.
From the introduction: "Let $H\sb k=\sum\sb {i=1}\sp k1/i$.
The Euler constant is defined as
$$C=\lim\sb {k\to\infty}\sum\sb {i=1}\sp k\left(\frac1i-\ln\frac{i+1}{i}\right)=
0.5772156649\cdots.$$ Hence $C=H\sb k-\ln(1+k)+\alpha\sb k$, where
$\alpha\sb k>0$, and then
(1) $H\sb k=\ln(1+k)+C+\gamma\sb k$, where $\gamma\sb k=-\alpha\sb k<0$.
"There exists for each integer $k>1$ a corresponding integer $n\sb k$ such
that $H\sb {n\sb k-1}= {vert}a{sub}2{vert} >=
{vert}a{sub}3{vert} >= ... and lim a{sub}n=0 holds.
Reviewed by Josef Antoni
80a:40002 40A05
Wadhwa, A. D.
Some convergent subseries of the harmonic series.
Amer. Math. Monthly 85 (1978), no. 8, 661--663.
It is known that if one omits the terms of the harmonic series {sum} 31/n
corresponding to the integers n whose decimal representations contain a
specified digit at least once, the resulting series converges. Let Z{sub}k
denote the set of nonnegative integers having exactly k zeros in their
decimal representation and let s{sub}k={sum}{sub}k 31/n{sub}k, where the
sum {sum}{sub}k is taken over n{sub}k{in}Z{sub}k. The author finds some
properties of the sequence (s{sub}k). The sequence (s{sub}k) is convex
and strictly decreasing. For every k, s{sub}k > 19.28. If {sum} epsilon {sub}k
is a convergent series of real terms, then {sum} 3 epsilon {sub}ks{sub}k is
convergent. The series {sum} 3(k+1) Delta {sup}2s{sub}k converges and
k Delta s{sub}k tends to zero. Let a{sub}0, 3a{sub}1, ... be positive numbers.
The sequence (t{sub}k), where t{sub}k={sum}{sub}k1/n{sup} alpha {sub}k,
converges for alpha > log{sub}1{sub}09 and decreases strictly for alpha >= 1.
The sequence (P{sub}k), where
P{sub}k={sum}{sub}k(a{sub}0n{sup}m{sub}k+a{sub}1n {sub}k{sup}(m - 1)+
...+a{sub}m){sup}( - 1), also decreases strictly.
The author shows that his result s{sub}k > 19.28 is equivalent to the
following theorem of R. P. Boas, Jr. ("Some remarkable sequences of integers",
to appear): The series {sum} 3a{sub}ks{sub}k converges or diverges according
as {sum} 3a{sub}k converges or diverges.
Reviewed by E. Reimers
55 #13119 40A25
Cerimele, B. J.
Summation of multiparameter harmonic series.
Fibonacci Quart. 15 (1977), no. 2, 116--117.
The author establishes two summation formulas for the so-called $\omega$-series, which enables
one to calculate the sums of $\omega$-series by means of some elementary functions. The
$\omega$-series is a series of the form $\omega(j;k\sb 1,\cdots,k\sb n)=\sum\sb {i=0}\sp
\infty(-1)\sp i/(j+s\sb i)$, where $j$ and $k\sb i (i=1,2,\cdots,n)$ are positive integers, $s\sb 0=0$,
$s\sb n=S$, $s\sb i=[i/n]S+\sum\sb {t=1}\sp {\text i,\text{mod}\,n}k\sb t$.
Reviewed by T. Salat
52 #1083 40A05
Wadhwa, A. D.
An interesting subseries of the harmonic series.
Amer. Math. Monthly 82 (1975), no. 9, 931--933.
If one omits from the harmonic series $\sum\sb {n=1}\sp \infty(1/n)$ the terms
corresponding to the integers $n$ whose decimal representation contains the
digit zero at least once, the resulting subseries converges
[A. J. Kempner, same Monthly 21 (1914), 48--50].
In this classroom note the sum of the thinned out series is shown to be
between 20.2 and 28.3. In fact, the series $\sum n\sp {-\alpha}$ converges
for $\alpha>\log\sb {10}9=0.95\cdots$, where the sum is over the
aforementioned integers. Corresponding results are obtained for integers
represented in other bases. All the proofs are elementary.
Reviewed by F. W. Hartmann
47 #2219 40A05
Demos, Miltiades S.
Class notes on series related to the harmonic series.
Math. Mag. 46 (1973), 40--41.
44 #7179 40.10
Boas, R. P., Jr.; Wrench, J. W., Jr.
Partial sums of the harmonic series.
Amer. Math. Monthly 78 1971 864--870.
The following theorem is established: Let $n\sb A$ denote the smallest
integer $n$ for which $\sum\sb {k=1}\sp n1/k$ exceeds $A(\geq 3)$; let
$e\sp {A-\gamma}=m+\delta$, where $m$ is an integer, $0<\delta<1$, and
$\gamma$ is Euler's constant; then, $n\sb A=m$ if
$\delta<{\textstyle\frac 1{2}}-(10n)\sp {-1}$, and $n\sb A=m+1$ if
$\delta>{\textstyle\frac 1{2}}+n\sp {-1}$. There are also numerical
tables giving $e\sp {A-\gamma}$ to 16 significant figures, $A=1$ through
20 and $A=100$, and values of the partial sums $S\sb n$ of $\sum 1/k$ to
10 decimal places for various values of $n$.
Reviewed by F. P. Cass
25 #2000 05.05
Gould, H. W.
Some relations involving the finite harmonic series.
Math. Mag. 34 1960/1961 317--321.
A unified method is given for the summation of certain finite series, involving initial segments of
the harmonic series and binomial coefficients. Examples are: $$ \gather n\sp 2\sum\sb {k=0}\sp
n(-1)\sp {n+k}[(n+k-1)!][(n-k)!k!k!]\sp {-1}\sum\sb {j=1}\sp {n+k-1}j\sp {-1}=1,\\ \sum\sb
{k=1}\sp n(-1)\sp {k-1}\left( \matrix n \\ k \endmatrix \right)\sum\sb {j=1}\sp {2k}j\sp
{-1}=(2n)\sp {-1}+2\sp {2n-2}(n-1)!/(2n-1)!,\\ \sum\sb {k=1}\sp n(-1)\sp {k-1}\left( \matrix n
\\ k \endmatrix \right)\left( \matrix 2k \\ k \endmatrix \right)\sp {-1}2\sp {2k}/k=2\sum\sb
{j=1}\sp {2n}j\sp {-1}-\sum\sb {j=1}\sp nj\sp {-1}. \endgather $$
Reviewed by Z. A. Melzak
7,413e 10.0X
Erd\Hos, Paul; Niven, Ivan
Some properties of partial sums of the harmonic series.
Bull. Amer. Math. Soc. 52, (1946). 248--251.
It was shown by the reviewer [T. Nagell, Skr. Norske Vid. Akad. Kristiania. I. 1923, no. 13,
10--15 (1924)] that $\sum\sb {k=0}\sp n(m+kd)\sp {-1}$ cannot be an integer. The authors prove
the following theorem of a similar nature. There is only a finite number of integers $n$ for which
one or more of the elementary symmetric functions of $1,{\textstyle\frac 1{2}},{\textstyle\frac
1{3}},\cdots,{\textstyle\frac 1{n}}$ is an integer. The proof is based on the result of A. E. Ingham
[Quart. J. Math., Oxford Ser. 8, 255--266 (1937)] that there is a prime between $x$ and $x+x\sp
{5/8}$. The authors assert, without proof, that the same result holds for the elementary symmetric
functions of $1/m,1/(m+1),\cdots,1/n$ and of $1/m,1/(m+d),1/(m+2d),\cdots,1/(m+nd)$. The
second result of this paper is the following theorem. No two partial sums of the harmonic series
can be equal; that is, it is not possible that $\sum\sb {k=m}\sp nk\sp {-1}=\sum\sb {k=x}\sp yk\sp
{-1}$. The proof depends on the application of the following theorem of I. Schur [cf. Erdos, J.
London Math. Soc. 9, 282--288 (1934)]. When $n>k$, there is in the set $n,n+1,\cdots,n+k-1$ at
least one integer containing a prime divisor greater than $k$.
Reviewed by T. Nagell
7,11i 40.0X
Erd\Hos, Paul; Niven, Ivan
On certain variations of the harmonic series.
Bull. Amer. Math. Soc. 51, (1945). 433--436.
Conditions are given for convergence of special series obtained by changing the signs of blocks of
terms of the harmonic series $\sum n\sp {-1}$. Let $u\sb p=p\sp {-1}$ and let $S(p,q)=u\sb
p+u\sb {p+1}+\cdots+u\sb {p+q-1}$. With $n$ and $k$ fixed positive integers, let $k\sb 2$ be the
greatest integer for which $S(n+k,k\sb 2)~~0$ and $T\sb p$ is decreasing. If $n$ and $k$ are such that $k\sb 2=k$, then $k\sb j=k$
when $j=3,4,5,\cdots$; in this case, each $T\sb p$ contains exactly $k$ terms of the harmonic
series, so that $T\sb p\rightarrow 0$ and (1) converges. If $n$ and $k$ are such that $k\sb 2\neq k$
(and hence $k\sb 2>k$), then the decreasing sequence $T\sb p$ has a positive limit and (1) diverges.
The proofs depend on elementary integral inequalities.
Reviewed by R. P. Agnew
5,117a 40.0X
Soddy, F.
The three infinite harmonic series and their sums (with topical reference to the Newton
and Leibniz series for $\pi$).
Proc. Roy. Soc. London. Ser. A. 182, (1943). 113--129.
The harmonic series, both with and without alternating signs, and their relationship with the series
obtained by continuing the progression backwards to $-\infty$ are discussed. A twelve place table
of sums of the harmonic series with alternating signs for various integral values of the difference
and first term of the associated arithmetic progression is included. The author's argument is
difficult to follow because of the formal viewpoint adopted. No apparent justification is given for
the necessary rearrangements of order of these conditionally convergent series. No reference is
given to Gauss, whose integral for the sum of the harmonic series with alternating signs seems to
include the author's results and more [see, for instance, Bromwich, Infinite Series, 2nd ed.,
Macmillan, London, 1926, p. 189].
Reviewed by P. W. Ketchum
3,295c 40.0X
Soddy, F.
The summation of infinite harmonic series.
Proc. Roy. Soc. London. Ser. A. 179, (1942). 377--386.
Harmonic series with alternating signs are transformed into more rapidly converging series. The
device to accomplish this result is based on the formal property of such series that, if any $n+1$
successive terms are multiplied by the binomial coefficients of the $n$th power of a binomial, the
result can be summed into one term of simple form, namely, a constant divided by the product of
the $n+1$ denominators of the terms. The terms of the given series are grouped into successive
groups of $n+1$ terms each, then multiplied by the binomial coefficients, and each group summed
by this principle. Finally an averaging process is performed over this sum and the similar sums
obtained by shifting all the groups a certain number of spaces to the right. The result is particularly
simple in case the series runs from $-\infty$ to $+\infty$. Various special cases are considered in
which very rapidly converging series are obtained for $\pi$, $\log\sb e2$ and other constants.
Reviewed by P. W. Ketchum
99e:11009 11A41
Gessel, Ira M.(1-BRND)
Wolstenholme revisited.
Amer. Math. Monthly 105 (1998), no. 7, 657--658.
In this short note, for positive integers $m$ and $k$, the author denotes by $R\sb m$ the set of
integers from $1$ to $m-1$ relatively prime to $m$, and $S(m,k)={\sum\sb {i\in{R\sb
m}}}1/i\sp k$. He proves, in a simple manner, the following: Theorem 1. If $k$ is not a multiple
of $p-1$ for any prime $p$ dividing $m$, then $S(m,k)\equiv 0$ (mod $m$). Theorem 2. If $k$ is
odd, and $k+1$ is not a multiple of $p-1$ for any prime $p$ dividing $m$, then $S(m,k)\equiv 0$
(mod ${m\sp 2}$).
Reviewed by Laurentiu Panaitopol
98e:11007 11A41
Bayat, M.
A generalization of Wolstenholme's theorem.
Amer. Math. Monthly 104 (1997), no. 6, 557--560.
The paper contains a proof of Wolstenhome's theorem, which states that the numerator of
$1+{1\over 2}+\cdots+1/(p-1)$ is divisible by $p\sp 2$ if $p>3$ is prime. The author states that
this result is true for $p\ge 3$; however, it fails to be correct for $p=3$. The same incorrect
condition appears also in Theorem 2 of the paper. In Theorem 3 it is stated that if $p$ is prime and
$k$ is a positive integer such that $2k\le p-1$, then the numerator of $$ 1+ {1\over 2\sp
{2k-1}}+\cdots+{1\over (p-1)\sp {2k-1}} $$ is divisible by $p\sp 2$. The correct condition is
$2k1\}$, and, moreover, it represents a holomorphic function in that region. Riemann
showed that the zeta function has an analytic continuation to the complex plane as a meromorphic
function which has only one singularity, a simple pole of residue 1 at the point 1. He also showed
that the zeta function satisfies a functional equation of the form $\zeta(1-s)=\gamma(s)\zeta(s)$,
$\gamma(s)=\pi\sp {1/2-s}\Gamma(s/2)œ\Gamma((1-s)/2)$, where $\Gamma$ denotes the gamma
function. In a certain sense Euler had already found the functional equation in 1749. There is
another representation of $\zeta$ due to Euler in 1749, the so-called product expansion
$\zeta(s)=\prod\sb p(1-p\sp {-s})\sp {-1}$ where the product is taken over all prime numbers $p$.
This is a fundamental expression and is the reason for the significance of the zeta function.
The author traces the development of the theory from Oresme to the work of Jakob and Johann
Bernoulli and looks at the research of Gottfried Wilhelm Leibniz and Leonard Euler. He discusses
the various approaches to evaluating $\zeta(2)$ and, more generally, $\zeta(2k)$, based on a
detailed study of primary source material. The decisive role played by the Poisson summation
formula is stressed. In view of, e.g., the historical treatise of H. M. Edwards [Riemann's zeta
function, Academic Press, New York, 1974; MR 57 #5922], no attempt is made to cover
Riemann's work and the subsequent development.
Reviewed by Joachim Schwermer
--
mailto:Torsten.Sillke@uni-bielefeld.de
http://www.mathematik.uni-bielefeld.de/~sillke/
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