$ \sum_{n=1}^\infty ( n ( \sum_{k=n}^\infty \frac{1}{k^2})^2 - \frac1n) = \frac32 - \frac12 \zeta(2) + \frac32\zeta(3)$ 
Prove  $$\sum_{n=1}^\infty \left( n \left( \sum_{k=n}^\infty \frac{1}{k^2}\right)^2 - \frac1n\right) = \frac32 - \frac12 \zeta(2) + \frac32\zeta(3)$$

I decided to write the LHS of the equation as
\begin{align}
\sum_{n=1}^\infty \left( n \left( \sum_{k=n}^\infty \frac{1}{k^2}\right)^2 - \frac1n\right) &= \sum_{n=1}^\infty n \left( \left( \sum_{k=n}^\infty \frac{1}{k^2}\right)^2 - \frac{1}{n^2}\right)\\
&= \sum_{n=1}^\infty n \left(\sum_{k=n}^\infty \frac{1}{k^2} - \frac{1}{n^2}\right)\left( \sum_{k=n}^\infty \frac{1}{k^2} + \frac{1}{n^2}\right)
\end{align}
I'm not sure how to proceed from here. Some hints would be greatly appreciated!

One way is to reduce the power of summand by summation by parts.
Take
$$a_n=\bigg(\sum_{k=n}^\infty\frac1{k^2}\bigg)^2-\frac1{n^2},\quad b_n=n,\quad B_n=\sum_{k=1}^nb_n=\frac12n(n+1),$$
we wish to find the closed form of
$$S=\sum_{n=1}^\infty a_nb_n.$$
By summation by parts, with some simplifications,
\begin{align*}
S&=\underbrace{\lim_{n\to\infty}a_nB_n}_{:=L=0}-
\sum_{n=1}^\infty B_n (a_{n+1}-a_n)\\
&=\sum_{n=1}^\infty \frac12\left(\frac1{n^2}+\frac1{n^3}+\frac1{n(n+1)}\right)
+\frac1n\sum_{k=n+1}^\infty\frac1{k^2}+\sum_{k=n+1}^\infty\frac1{k^2}-\frac1n\\
&=\frac12(\zeta(2)+\zeta(3)+1)
+\underbrace{\sum_{n=1}^\infty\sum_{k=n+1}^\infty\frac1{nk^2}}_{:=S'=\zeta(3)}
+\underbrace{\sum_{n=1}^\infty\bigg(\sum_{k=n+1}^\infty\frac1{k^2}-\frac1n\bigg)}_{:=S''=1-\zeta(2)}.
\end{align*}


*

*Proof of $L=0$: By Stolz–Cesàro,
$$\lim_{n\to\infty}\frac{\sum_{k=n}^\infty\frac1{k^2}}{\frac1n}
=\lim_{n\to\infty}\frac{\frac1{n^2}}{\frac1{n(n+1)}}=1\quad 
\implies\quad \sum_{k=n}^\infty\frac1{k^2}=\frac1n+o\left(\frac1n\right).$$
Thus,
$$\lim_{n\to\infty}a_nB_n=\lim_{n\to\infty}o\left(\frac1{n^2}\right)\frac{n(n+1)}2=0.$$


*Proof of $S'=\zeta(3)$:
\begin{align*}
\sum_{n=1}^\infty\sum_{k=n+1}^\infty\frac1{nk^2}
&=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n(n+k)^2}\\
&=\frac12\sum_{n=1}^\infty\sum_{k=1}^\infty\left(\frac1n+\frac1k\right)\frac1{(n+k)^2}\\
&=\frac12\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{nk}\int_0^1 x^{n+k-1}\,\mathrm dx\\
&=\frac12\int_0^1\frac1x\sum_{n=1}^\infty\frac{x^n}n\sum_{k=1}^\infty\frac{x^k}k\,\mathrm dx\\
&=\frac12\int_0^1\frac{\ln^2(1-x)}x\,\mathrm dx\\
&=\zeta(3).
\end{align*}


*Proof of $S''=1-\zeta(2)$: We consider the partial sum. Summation by parts gives
$$\sum_{n=1}^\ell\sum_{k=n+1}^\infty\frac1{k^2}
=\ell\sum_{n=\ell+1}^\infty\frac1{k^2}
+\sum_{n=1}^{\ell-1}\frac n{(n+1)^2}.$$
As $\ell\to\infty$,
$$\sum_{n=1}^\infty\bigg(\sum_{k=n+1}^\infty\frac1{k^2}-\frac1n\bigg)
=\underbrace{\lim_{\ell\to\infty}\ell\sum_{n=\ell+1}^\infty\frac1{k^2}}_{=1}
+\underbrace{\sum_{n=1}^{\infty}\bigg(\frac n{(n+1)^2}-\frac1n\bigg)}_{=-\zeta(2)}.$$
