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.
 A: Following Leucipus' answer, I will use different approach to evaluate $S(1)$. Note
\begin{eqnarray}
S=S(1)&=&\int_0^1\int_0^xS''(x)dxt\\
&=&-\int_0^1\int_0^x\frac{\ln(1-x^2)}{1-x^2}dxdt\\
&=&-\int_0^1\int_x^1\frac{\ln(1-x^2)}{1-x^2}dtdx\\
&=&-\int_0^1(1-x)\frac{\ln(1-x^2)}{1-x^2}dx\\
&=&-\int_0^1\frac{\ln(1-x^2)}{1+x}dx\\
&=&-\int_0^1\frac{\ln(1-x)}{1+x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx\\
&=:&-I_1-I_2
\end{eqnarray}
Now
$$ I_2=\int_0^1\ln(1+x)d\ln(1+x)=\frac{1}{2}\ln^2(1+x)\bigg|_0^1=\frac12\ln^22. $$
For $I_1$, under $t=1-x$, one has
\begin{eqnarray}
\int\frac{\ln t}{2-t}dt&=&-\int_0^1\ln xd\ln(2-t)\\
&=&-\ln x\ln(2-t)+\int\frac{\ln(2-t)}{t}dt\\
&=&-\ln x\ln(2-t)+\int\frac{\ln2+\ln(1-\frac12t)}{t}dt\\
&=&-\ln x\ln(2-t)+\ln2\ln t+Li_2(\frac t2)+C
\end{eqnarray}
\begin{eqnarray}
I_1&=&\int_0^1\frac{\ln t}{2-t}dt=-\int_0^1\ln xd\ln(2-t)\\
&=&-\ln x\ln(2-t)+\ln2\ln t+Li_2(\frac t2)\bigg|_0^1
&=&-\frac{\pi^2}{12}+\frac12\ln^22.
\end{eqnarray}
Thus
$$ S=\frac{\pi^2}{12}-\ln^22. $$
A: 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}
A: A different view on the same problem:
$$ \sum_{n\geq 1}\frac{x^n}{n}=-\log(1-x),\qquad \sum_{n\geq 1}H_n x^n = \frac{-\log(1-x)}{1-x} \tag{A}$$
$$ \sum_{n\geq 1}H_n x^{2n} = \frac{-\log(1-x^2)}{1-x^2}\tag{B}$$
$$\begin{eqnarray*}\sum_{n\geq 1}H_n\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right) &=& \int_{0}^{1}\frac{-\log(1-x^2)}{1+x}\,dx\\&=&-\tfrac{1}{2}\log^22+\int_{0}^{1}\frac{-\log(1-x)}{1+x}\,dx\\&=&-\tfrac{1}{2}\log^22+\int_{0}^{1}\frac{-\log(x)}{2-x}\,dx\tag{C}\end{eqnarray*} $$
and by differentiation under the integral sign, the last integral is related to the series
$$ \sum_{n\geq 1}\frac{1}{n^2 2^n}=\text{Li}_2\left(\tfrac{1}{2}\right)\stackrel{(*)}{=}\tfrac{\pi^2}{12}-\tfrac{\log^2 2}{2} \tag{D}$$
where $(*)$ follows from the dilogarithm reflection formula, proved here.
A: 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
$$S(1) = \sum_{n=1}^{\infty}\frac{H_{n}}{(2n+1)(2n+2)} = \frac{\pi^2}{12} - \ln^{2}2$$
A: A slight variant on other answers: we want to evaluate$$\sum_{n\ge0}\int_0^1H_nx^{2n}(1-x)dx=\int_0^1\frac{\ln(1-x^2)}{1+x}dx\\=\left[\frac12\ln^2(1+x)+\operatorname{Li}_2\left(\frac{1-x}{2}\right)+\ln(1-x)\ln\frac{1+x}{2}\right]_0^1\\=\frac12\ln^22-\operatorname{Li}_2\left(\frac12\right)+\lim_{x\to1^-}\ln(1-x)\ln\frac{1+x}{2}=\ln^22-\frac{\pi^2}{12}.$$In particular, the limit is $0$, since $$\lim_{y\to0^+}\ln y\ln\left(1-\frac{y}{2}\right)=-\frac12\lim_{y\to0^+}y\ln y=0.$$
A: Different approach
$$\sum_{k=1}^\infty\frac{H_k}{(2k+1)(2k+2)}=\frac14\sum_{k=1}^\infty\frac{H_k}{(k+1/2)(k+1)}$$
By the master theorem we have 
$$\sum_{k=1}^\infty\frac{H_k}{(k+n+1)(k+1)}=\frac{H_n^2+H_n^{(2)}}{2n}$$
By setting $n=-1/2$ , the result follows. Note that we used  $H_{-1/2}=-2\ln2$ and $H_{-1/2}^{(2)}=-2\zeta(2)$
