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A294970 Numerators of the partial sums for the Catalan constant A006752: Sum_{k=0..n} ((-1)^k)/(2*k+1)^2, n >= 0. 4

%I #16 Sep 08 2022 08:46:20

%S 1,8,209,10016,91369,10956424,1863641881,1854623872,538015351033,

%T 193637145687688,194117166024913,102476291858462752,

%U 2566386635039604121,23062916917686411464,19421109407275720721849,18642496069331249273291264

%N Numerators of the partial sums for the Catalan constant A006752: Sum_{k=0..n} ((-1)^k)/(2*k+1)^2, n >= 0.

%C The corresponding denominators are given in A294971.

%H G. C. Greubel, <a href="/A294970/b294970.txt">Table of n, a(n) for n = 0..575</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HurwitzZetaFunction.html">Hurwitz Zeta Function</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TrigammaFunction.html">Trigamma Function</a>

%F a(n) = numerator(r(n)) with the rationals r(n) = Sum_{k=0..n} (-1)^k/(2*k+1)^2 = (Zeta(2, 1/4) - Zeta(2,floor(n/2) + 5/4) - (Zeta(2, 3/4) - Zeta(2,floor((n-1)/2) + 7/4)))/16, Zeta(2, z) = Psi(1, z), with the Hurwitz Zeta function and the trigamma function Psi(1, z).

%F The limit n-> infinity of r(n) is the Catalan constant given in A006752; see in particular the formula (Zeta(2, 1/4) - (Zeta(2, 3/4))/16.

%e The rationals r(n) begin: 1, 8/9, 209/225, 10016/11025, 91369/99225, 10956424/12006225, 1863641881/2029052025, 1854623872/2029052025, 538015351033/586396035225, 193637145687688/211688968716225, 194117166024913/211688968716225, 102476291858462752/111983464450883025, ...

%e r(10^5) = 0.9159655942 (Maple 10 digits) to be compares with 0.91596559417... from A006752.

%t Table[Numerator[Sum[(-1)^k/(2*k+1)^2, {k,0,n}]], {n,0,20}] (* _Vaclav Kotesovec_, Nov 15 2017 *)

%o (PARI) for(n=0,20, print1(numerator(sum(k=0,n, (-1)^k/(2*k+1)^2)), ", ")) \\ _G. C. Greubel_, Aug 22 2018

%o (Magma) [Numerator((&+[(-1)^k/(2*k+1)^2: k in [0..n]])): n in [0..20]]; // _G. C. Greubel_, Aug 22 2018

%Y Cf. A006752, A294971.

%K nonn,frac,easy

%O 0,2

%A _Wolfdieter Lang_, Nov 15 2017

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