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Finding $$\sum^{\infty}_{n=1}\bigg[\bigg(\frac{(-1)^{n+1}}{n+1}\bigg)\bigg(\sum^{n}_{r=1}\frac{1}{r}\bigg)\bigg]$$

Try: Let $\displaystyle H_{n} =\sum^{n}_{r=1}\frac{1}{r},$ then series $\displaystyle \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n+1}\cdot H_{n}=\frac{H_{1}}{2}-\frac{H_{2}}{3}+\frac{H_{3}}{4}\cdots\cdots$

could some help me to solve it, thanks

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My solution (sorry, didn't have time to write it up earlier) isn't that short, but rather straightforward: I use the integral representations $$H_n=\int^1_0\frac{1-t^n}{1-t}\,dt$$ and $$\frac1{n+1}=\int^1_0u^n\,du,$$ so we have \begin{align}\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n+1}\,H_{n}&=\int^1_0\int^1_0\sum^\infty_{n=1}(-1)^{n+1}\frac{u^n-u^nt^n}{1-t}\,dt\,du\\&=\int^1_0\int^1_0\left(\frac{u}{1+u}-\frac{ut}{1+ut}\right)\frac1{1-t}\,dt\,du\\&=\int^1_0\int^1_0\frac{u}{(1+u)(1+ut)}\,dt\,du\\&=\int^1_0\frac{\ln(1+u)}{1+u}\,du=\frac12\,\ln^22\end{align}

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Summation by parts is a good way to go, generating functions another one. We have $$ -\log(1-x) = \sum_{n\geq 1}\frac{x^n}{n}\tag{1} $$ $$ \frac{-\log(1-x)}{1-x} = \sum_{n\geq 1} H_n\,x^n\tag{2} $$ $$\sum_{n\geq 1}\frac{H_n}{n+1}\,z^{n+1} = \int_{0}^{z}\frac{-\log(1-x)}{1-x}\,dx = \frac{1}{2}\log^2(1-z)\tag{3}$$ and by applying $\lim_{z\to -1^+}$ to both sides of $(3)$ you may easily recover Professor Vector's claim in the comments.

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  • $\begingroup$ Nice Jack , please explain me second line ,how can i get it, thanks $\endgroup$ – DXT Dec 28 '17 at 5:08
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n + 1}\,H_{n} & = \sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,H_{n}\int_{0}^{1}t^{n}\,\dd t = -\int_{0}^{1}\sum_{n = 1}^{\infty}H_{n}\pars{-t}^{n}\,\dd t = \int_{0}^{1}{\ln\pars{1 + t} \over 1 + t}\,\dd t \\[5mm] & = \left.{1 \over 2}\,\ln^{2}\pars{1 + t}\,\right\vert_{\ 0}^{1} = \bbx{{1 \over 2}\,\ln^{2}\pars{2}} \end{align}

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