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Evaluate: $$\sum_{k=1}^\infty\left(\zeta(2)-\sum_{n=1}^k\frac1{n^2}\right)^2$$

Recognizing that $\zeta(2)-\sum_{n=1}^k\frac1{n^2}$ can be written as $\psi_1(1+k)$ where $\psi_1(z)$ is the trigamma function, What remains to be done is to evaluate: $$\sum_{k=1}^\infty\psi_1^2(k+1)$$ Mathematica could not evaluate it in a closed form but the source assures that it exists.

If you liked this problem check out Hard Definite integral involving the Zeta function.

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Divyansh Garg
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  • @DavidH Please don't mark this question as a duplicate as I would like to see other ingenious solutions to the problem – Divyansh Garg Aug 29 '14 at 11:22
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    Your sum evaluates to $3\zeta(3)-\zeta(2)^2$. – David H Aug 29 '14 at 11:25
  • @DavidH, it seems to me that the proposed duplicate talks about the derivative of the trigamma function $\psi'$, not about $\psi$ itself. – Ian Mateus Aug 29 '14 at 11:37
  • @IanMateus I think you're confused. $\psi$ represents the digamma function, not the trigamma function. The duplicate I proposed talks about the derivative the digamma function $\psi^\prime$, which is the trigamma function. – David H Aug 29 '14 at 12:22
  • @DavidH I see, thanks! It is indeed a duplicate, then. – Ian Mateus Aug 29 '14 at 12:29
  • I cannot post an answer but I found that the result is $3\zeta(3)-\frac52\zeta(4)$ using standard summation techniques after writing the squared term as a series and expanding into two series. – Tom-Tom Aug 29 '14 at 13:13
  • @DivyanshGarg: While your question is a duplicate, seeing it lead me to the other question, to which I have added an answer. I hope that satisfies some of your desire to see other answers. (+1) – robjohn Aug 29 '14 at 19:12

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Using Maple I am obtaining

$$\sum _{k=1}^{\infty } \left( \Psi \left( 1,k+1 \right) \right) ^{2}= 0.9003626252$$

Juan Ospina
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