2
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Inspired by the identity listed in this problem, I tried

2 π Sqrt[2]/9801 Sum[((4 k)! (1103 + 36390 k))/((k!)^4  396^(4 k)), 
  {k, 0, \[Infinity]}]

and got

(1/(418435621956 Sqrt[2])) π 
(188362204272 HypergeometricPFQ[{1/4, 1/2, 3/4}, {1, 1}, 1/96059601] + 
6065 HypergeometricPFQ[{5/4, 3/2, 7/4}, {2, 2}, 1/96059601])

FullSimplify did not lead to any simplification or the desired answer: $1$.

However, N[%] yielded 1.

Why didn't this summation equality get evaluated symbolically to yield the answer $1$?

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  • $\begingroup$ In fact, N[%-1] doesn't yield zero, it yields $8.8\cdot 10^{-9}$. Are you completely sure the sum is indeed one? Or may it just be close? $\endgroup$ – AccidentalFourierTransform Feb 3 '18 at 0:37
  • $\begingroup$ I'm assuming the equality identity (not approximation) given in the linked problem is correct, but perhaps this is a mistake. $\endgroup$ – David G. Stork Feb 3 '18 at 0:39
  • $\begingroup$ I got the one when I run this and I ask for the numerical value. However, when I subtract one from the other I get something close to zero, namely $-2.81647*10^{-9}$. $\endgroup$ – Darth_Bane Feb 3 '18 at 0:42
6
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The correct formula is with a 26390 instead of 36390. See (78) here. Mathematica now outputs the correct result:

\[Pi] Sqrt[8]/9801 Sum[((4 k)! (1103 + 26390 k))/((k!)^4 396^(4 k)), {k, 0, \[Infinity]}]
(* 1 *)

Hooray!

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  • $\begingroup$ Excellent! Thanks! $\endgroup$ – David G. Stork Feb 3 '18 at 0:56

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