Strategies For Summing Harmonic Numbers

You may evaluate it without using generating functions.

$$S=\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^2} = \sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{n}\frac{1}{n^2k^2}$$

By changing the order of summation, you may write it as:

$$\begin{align}S &= \sum\limits_{k=1}^{\infty}\sum\limits_{n=k}^{\infty}\frac{1}{n^2k^2}\\&= \sum\limits_{k=1}^{\infty}\frac{1}{k^2}\left(\sum\limits_{n=k}^{\infty}\frac{1}{n^2}\right)\\&= \sum\limits_{k=1}^{\infty}\frac{1}{k^2}\left(\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - \sum\limits_{n=1}^{k-1}\frac{1}{n^2}\right) \\&= \sum\limits_{k=1}^{\infty}\frac{1}{k^2}\left(\zeta(2) - \sum\limits_{n=1}^{k-1}\frac{1}{n^2}\right)\\&= \sum\limits_{k=1}^{\infty}\frac{1}{k^2}\left(\zeta(2) + \frac{1}{k^2} - H_k^{(2)}\right)\\&= \zeta(2)\sum\limits_{k=1}^{\infty}\frac{1}{k^2} + \sum\limits_{k=1}^{\infty}\frac{1}{k^4} - \sum\limits_{k=1}^{\infty}\frac{H_k^{(2)}}{k^2} \\&= \zeta^2(2)+\zeta(4) - S\end{align}$$

Hence, $\displaystyle S = \frac{\zeta^2(2)+\zeta(4)}{2}$.

In general for any sequence $(a_n)_{n \ge 1}$ such that $\displaystyle \sum_{n=1}^{\infty} a_n$ converges absolutely,

We have $$\sum_{n=1}^{\infty}\left(a_n\sum_{k=1}^{n}a_k\right) = \frac{1}{2}\left(\sum_{n=1}^{\infty}a_n^2 + \left(\sum_{n=1}^{\infty}a_n\right)^2\right)$$ by using the same method as above.

Here is a general version I discovered some time ago.

Proposition : $$\sum_{r=1}^{n} \dfrac{H_{r} ^{(m)}}{r^m} = \dfrac{1}{2} \left( [H_{n}^{(m)}]^2 + H_{n}^{(2m)} \right) \quad ; \quad m \geq 1$$

Proof : Expand the summation, using the definition of $\displaystyle H_{r}^{(m)}$ , to see that $$\displaystyle \sum_{r=1}^{n} \dfrac{H_{r}^{(m)}}{r^m} = \sum_{r=1}^{n} \dfrac{1}{r^{2m}} + \sum_{r < j} \sum_{j=2}^{n} \dfrac{1}{rj} = H_{n}^{(2m)} + \sum_{r < j} \sum_{j=2}^{n} \dfrac{1}{rj} $$

And,

$\displaystyle [H_{n}^{(m)}]^2 = H_{n}^{(2m)} + 2\sum_{r < j} \sum_{j=2}^{n} \dfrac{1}{rj} $

Eliminating $\displaystyle \sum_{r < j} \sum_{j=2}^{n} \dfrac{1}{rj}$ from the above equations, we have,

$$\displaystyle \sum_{r=1}^{n} \dfrac{H_{r}^{(m)}}{r^m} =\dfrac{1}{2} \left( [H_{n}^{(m)}]^2 + H_{n}^{(2m)} \right) \quad \square \tag{*} $$

Now put $m=2$ take limit to infinity to get the desired result.

We can also derive the result using Summation By Parts.