<s>[INST] How to evaluate the sum $\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$ How to evaluate the sum: $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$$
Can anyone help me,I really appreciate it.
 [/INST] It is evident that
$$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right) = \frac{\pi^2}{12} - \ln^{2}2$$
and can be evaluated by following the pattern:
Consider the series
$$S(x) = \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+2}}{(2n+1)(2n+2)}$$
which upon differentiation leads to $S(0) = 0$, $S'(0) = 0$, 
\begin{align}
S(x) &= \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+2}}{(2n+1)(2n+2)} \\
S'(x) &= \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+1}}{(2n+1)} \\
S''(x) &= \sum_{n=1}^{\infty} H_{n} \, x^{2n} = - \frac{\ln(1- x^2)}{1-x^2}.
\end{align}
Now,
$$ - 2 \, S''(x) = \frac{\ln(1-x)}{1-x} + \frac{\ln(1-x)}{1+x} + \frac{\ln(1+x)}{1-x} + \frac{\ln(1+x)}{1+x}$$
which, upon integration, leads to
\begin{align}
- 4 \, S'(x) &= \ln^{2}(1 + x) - \ln^{2}(1-x) + 2 \, Li_{2}\left(\frac{1-x}{2}\right) - 2 \, Li_{2}\left(\frac{1+x}{2}\right) + \ln4 \, \ln\left(\frac{1+x}{1-x}\right).
\end{align}
Integrating again leads to $S(x)$. The integrals
\begin{align}
\int_{0}^{x} \ln^{2}(1-t) \, dt &= (x-1) \, (\ln^{2}(1-x) - 2 \ln(1-x) + 2) + 2 \\
\int_{0}^{x} \ln^{2}(1+t) \, dt &= (x+1) \, (\ln^{2}(1+x) - 2 \ln(1+x) + 2) - 2 \\ 
\int_{0}^{x} \ln\left(\frac{1+t}{1-t}\right) \, dt &= x \, \ln\left(\frac{1+x}{1-x}\right) + \ln(1-x^2) \\
\int_{0}^{x} Li_{2}\left(\frac{1+t}{2}\right) \, dt &= (1+x) \, Li_{2}\left(\frac{1+x}{2}\right) + x \, \ln\left(\frac{1-x}{2}\right) - \ln(1-x) -x - Li_{2}\left(\frac{1}{2}\right) \\
\int_{0}^{x} Li_{2}\left(\frac{1-t}{2}\right) \, dt &= (x-1) \, Li_{2}\left(\frac{1-x}{2}\right) + (x+1) \, \ln\left(\frac{1+x}{2}\right) -x - Li_{2}\left(\frac{1}{2}\right) + \ln2
\end{align} 
are needed for the evaluation. Once $S(x)$ is determined set $x=1$ to obtain
$$S(1) = \sum_{n=1}^{\infty}\frac{H_{n}}{(2n+1)(2n+2)} = \frac{\pi^2}{12} - \ln^{2}2$$
</s>