Only zeta this time

We have, by Abel's summation formula,that $\sum_{n=1}^{\infty}\left(\zeta(2)- 1 - \frac{1}{2^2} - \dots - \frac{1}{n^2}\right)^2= \\2\sum _{ n=1 }^{ \infty }{ \frac { n }{ (n+1)^{ 2 } } \left( \zeta (2)-{ H }_{ n }^{ (2) } \right) } -\sum _{ n=1 }^{ \infty }{ \frac { n }{ (n+1)^{ 4 } } }=\\2\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n+1 } \left( \zeta (2)-{ H }_{ n }^{ (2) } \right) } -2\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (n+1)^{ 2 } } \left( \zeta (2)-{ H }_{ n }^{ (2) } \right) } -\zeta (3)+\zeta (4)$

in order to find the value of the first sum,we need the following lemma

Lemma : $\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n } \left( \zeta (2)-{ H }_{ n }^{ (2) } \right) } =\zeta (3)$

Proof :

We apply Abel's summation formula and we get that $\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n } \left( \zeta (2)-{ H }_{ n }^{ (2) } \right) } =\lim _{ n\rightarrow \infty }{ { H }_{ n }\left( \zeta (2)-{ H }_{ n+1 }^{ (2) } \right) } +\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n } }{ { (n+1) }^{ 2 } } } =\\ \sum _{ n=1 }^{ \infty }{ \left( \frac { { H }_{ n+1 } }{ { (n+1) }^{ 2 } } -\frac { 1 }{ (n+1)^{ 3 } } \right) } =\zeta (3)$

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we used that $\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n } }{ { n }^{ 2 } } } =2\zeta (3)$

the proof is simple and left as an exercise for the reader

Hint: $\int _{ 0 }^{ 1 }{ { x }^{ n }\ln { (1-x) } } =-\frac { { H }_{ n+1 } }{ n+1 }$

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back to our solution we have

$\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n+1 } \left( \zeta (2)-{ H }_{ n }^{ (2) } \right) }=\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n+1 } \left( \zeta (2)-{ H }_{ n+1 }^{ (2) } \right) } +\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (n+1)^{ 3 } } }$

using the lemma we get that the sum is equal to $2\zeta (3)-\zeta (2)$

on the other hand,a calculation shows that $\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ (n+1)^{ 2 } } \left( \zeta (2)-{ H }_{ n }^{ (2) } \right) } =-\int _{ 0 }^{ 1 }{ \frac { \ln { x } }{ x(1-x) } \left(\text{Li}_2(x) -x \right) }$

evaluating the integral we get $\frac { 7 }{ 4 } \zeta (4)-\zeta (2)$

putting all of those together we get the desired sum $\sum_{n=1}^{\infty}\left(\zeta(2)- 1 - \frac{1}{2^2} - \dots - \frac{1}{n^2}\right)^2= \Large\boxed{3\zeta (3)-\frac { 5 }{ 2 } \zeta (4)}$

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Notation used: ${ H }_{ n }^{ (s) }=\sum _{ k=1 }^{ n }{ \frac { 1 }{ { k }^{ s } } }$