<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] We have  $$\sum_{n=1}^\infty H_n x^{2n}=-\frac{\ln(1-x^2)}{1-x^2}$$
integrate both sides from $x=0$ to $x=y$, we get
$$\sum_{n=1}^\infty \frac{H_n y^{2n}}{2n+1}=-\int_0^y\frac{\ln(1-x^2)}{1-x^2}dy$$
Now integrate both sides from $y=0$ to $y=1$, we get
\begin{align}
S&=\sum_{n=1}^\infty \frac{H_n}{(2n+1)(2n+2)}=-\int_0^1\int_0^y\frac{\ln(1-x^2)}{1-x^2}\ dx\ dy\\
&=-\int_0^1\frac{\ln(1-x^2)}{1-x^2}\left(\int_x^1 dy\right)\ dx=-\int_0^1\frac{\ln(1-x^2)}{1-x^2}\left(1-x\right)\ dx\\
&=-\int_0^1\frac{\ln(1-x^2)}{1+x}\ dx\overset{\large\frac{1-x}{1-x}\mapsto x}{=}-\int_0^1\frac{2\ln2+\ln x-2\ln(1+x)}{1+x}\ dx\\
&=-2\ln^22-\int_0^1\frac{\ln x}{1+x}\ dx+2\int_0^1\frac{\ln(1+x)}{1+x}\ dx\\
&=-2\ln^22-\left(-\frac12\zeta(2)\right)+\ln^22\\
&\boxed{=\frac12\zeta(2)-\ln^22}
\end{align}
</s>