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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.

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  • $\begingroup$ is there a source for this sum? Or, just a practice problem? $\endgroup$ – karakfa Oct 20 '17 at 15:05
  • $\begingroup$ @MarkViola Mathematica gives the sum of the series. I don't know what is the trick to evaluate it. I think this is a good problem. :) $\endgroup$ – Ixion Oct 20 '17 at 15:26
  • $\begingroup$ Partial fraction decomposition $\endgroup$ – JohnColtraneisJC Oct 20 '17 at 15:27
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    $\begingroup$ Since $\displaystyle\frac {\pi^2}{12}=\sum_{r=1}^\infty \frac 1{2r^2}$, and $\displaystyle\ln 2=\sum_{r=1}^\infty \frac {(-1)^{r+1}}{r}$ is it possible to transform the given summation $\displaystyle \sum_{n=1}^\infty \sum_{r=1}^n \frac 1{(2n+1)(2n+2)r}$into $\displaystyle\sum_{r=1}^\infty \frac 1{2r^2}-\left(\sum_{r=1}^\infty \frac {(-1)^{r+1}}{r}\right)^2$? If so, then we are done. $\endgroup$ – hypergeometric Oct 21 '17 at 15:35
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    $\begingroup$ I am voting for reopening this question since, despite a lack of efforts from the OP, it led to a number of collateral questions, and I think it is correct to leave this original question as a reference. $\endgroup$ – Jack D'Aurizio Oct 26 '17 at 15:57
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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.

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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 $$S(1) = \sum_{n=1}^{\infty}\frac{H_{n}}{(2n+1)(2n+2)} = \frac{\pi^2}{12} - \ln^{2}2$$

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    $\begingroup$ Nice derivation. (+1) Nevertheless some lines are needed to make the evidence evident. :-) $\endgroup$ – Markus Scheuer Oct 20 '17 at 20:12
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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 $$ S=\frac{\pi^2}{12}-\ln^22. $$

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