Comments on A095820. The rational numbers r(n):= sum(1/k^n,k=1..n-1) + 1/((n-1)*(n-1)!), n>=2 (written in lowest terms, i.e. without common factors >1 in the numerator and denominator) are, for 2 .. 15: [2, 11/8, 1465/1296, 260467/248832, 47541136609/46656000000, 941124897061/933120000000, 972240507397068973121/968265199641600000000, 7727206375538178489426059/7711694390034432000000000, 10338017533904483647451374351534201/10327742657402407212810240000000000, 26038773922578490153470593775940352227/26025911496654066176281804800000000000, 205885410745914874422268554962004690219615313153956577/ 205834757111243649981930784496178708873216000000000000, 5706439638506708531123830750724622817764876710454626598761/ 5705739467123673977499121346234073809965547520000000000000, 124557157881364038235127115252745622586124979436486684496540198584405405227957/ 124549529453424021258327755672649951522982868564582695048650424320000000000000, 224420206577703990359490919696180285507535234365687562058382523333975596342707747869/ 224413342169179401503254950170980682654110532579665099938658334539776000000000000000] Their values (maple9 with 20 digits) are [2., 1.3750000000000000000, 1.1304012345679012346, 1.0467584555041152263, 1.0189715494041495199, 1.0085786362536436900, 1.0041055980912465092, 1.0020114886196465612, 1.0009948811509852949, 1.0004942161555447382, 1.0002460888305848590, 1.0001227135215459667, 1.0000612481474116662, 1.0000305882371263557] ############################################################################################ The real numbers Zeta(n):=sum(1/k^n,k=1..infty), n integer, n>=2, (Euler's, when generalized to complex n with Re(n)>1 called Riemann's, Zeta function) have the upper bounds r(n). The errors 0<= r(n)- Zeta(n) are (maple9, 20 digits), for n=2..15: [.3550659331517735635, .1729430968404057146, 0.480780008567630430e-1, 0.98307003607453000e-2, 0.16284874197003801e-2, 0.2293588717208632e-3, 0.282418933021697e-4, 0.30957935643468e-5, 0.3060231672095e-6, 0.275514252736e-7, 0.22772768105e-8, 0.1739674776e-9, 0.123529612e-10, 0.8193352e-12, 0.509808e-13, 0.29871e-14, 0.1652e-15, 0.87e-17, 0.2e-18] Here e-k at the end of a decimal means 10^{-k}. ##################################################################################################### A proof of this (trivial) inequality Zeta(n) <= r(n):= sum(1/k^n,k=1..n-1) + 1/((n-1)*(n-1)!), for n integer, n>=2, runs as follows. The following item i) is not necessary for this but included here for convenience. 1) Prove first that Zeta(n) converges absolutely for all real n with n>1. This can be done for Zeta(x) with real x>1, and is a standard proof in Analysis (e.g. H. Heuser: Analysis I, 15. Auflage, Teubner, 2003, Aufgabe 4, p.195, Loesung: p.599): Show that the partial sums s_n(x) are monotonly increasing (obvious) and bounded. For the boundedness one uses k with 2^k >n, i.e. n<= 2^k-1, and then s_n(x) <= s_{2^k-1} = 1 + (1/2^x + 1/3^x) + ... + (1/(2^{k-1})^x + ...+1/(2^k -1)^x) < 1 + 2/2^x = ... + (2^{k-1})/(2^{k-1})^x = (1 - 1/2^{x-1}^k)/(1 - 1/2^{x-1}) < 1/(1 - 1/2^{x-1}) =: K(x) (upper bound for x>1, n independent). 2) Use the trivial inequality 1/k^n < 1/(k*(k-1)*...*(k-(n-1))) =:A(n,k) , n and k integers, k>=n>=2. The family of infinite series s(n):= sum(A(n,k),k=n..infty), n>=2, can be computed as the limit of the partial sums s(n,N):= sum(A(n,k),k=n..N) for N to infinity. This is done by rewriting, for k>=n>=2, A(n,k) = (1/(k*(k-1)*...*(k-(n-2)) - 1/((k-1)*...*(k-(n-1))*(-1/(n-1)) =: B(n,k)-B(n,k-1) . Now s(n,N) becomes a telescope-sum with the result s(n,N) = B(n,n-1) + B(n,N). The second term vanishes in the limit of N to infinity (product of sequences with tend to 0). Therefore, s(n) = B(n,n-1) = 1/((n-1)*(n-1)!), n>=2. 3) Due to the inequality in 2), the convergent series z(n):= sum(1/k^n,k=n..infty) has upper bound s(n). Because Zeta(n) = sum(1/k^n, k=1..n-1) + z(n), and obviously Zeta(n)>0, one finally has 0 < Zeta(n)- sum(1/k^n, k=1..n-1) <= 1/((n-1)*(n-1)!), n integer, n>=2, which means that 0 < Zeta(n) <= r(n). ####################################################################################################### ########################################################################################################