Calculate $ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1} H_n$ Define
$$
H_n = \sum_{k=1}^n \frac{1}{k}
$$
I need to calculate the sum
$$
S = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1} H_n
$$
Using the following integral representation of $ H_n$
$$
H_n = -n \int_0^1 x^{n-1}\ln(1-x) dx
$$
and exchanging the order of summation, I obtained
$$
S = -\int_0^1 \left(\frac{1}{1+x}+\frac{\ln(1+x)-x}{x^2}\right) \ln(1-x) dx
$$
Using Wolfram Alpha, I got
$$
S \approx 0.240227
$$
so I guess
$$
S = \frac{(\ln 2)^2}{2}
$$
But I don't know how to calculate the integral. Any idea?
 A: 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}
A: 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.
A: Integrate by parts
\begin{align}
S =& -\int_0^1 \left(\frac{1}{1+x}+\frac{\ln(1+x)-x}{x^2}\right) \ln(1-x)\>dx\\
=& \int_0^1 d\left(\frac{\ln(1+x)}{x}-\ln2\right) \ln(1-x)
=\int_0^1 \frac{\ln(1+x)-x \ln2}{x(1-x)}dx \\
=&\int_0^1 \frac{\ln(1+x)}{x}dx + \int_0^1 \frac{\ln\frac{1+x}2}{1-x}dx\\
\end{align}
and then substitute $x\to \frac{1-x}{1+x}$ in the second integral
\begin{align}
S=\int_0^1 \frac{\ln (1+x)}{x}dx - \int_0^1 \frac{\ln({1+x})}{x(1+x)}dx
=\int_0^1 \frac{\ln(1+x)}{1+x}dx=\frac12 \ln^22
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
A: I thought it might be instructive to present a way forward the relies on using the series definition of $H_n$ and proceeding without using integrals.  To that end we proceed.

The series of interest is given by $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}H_n$ where we define $H_n$ as $H_n =\sum_{k=1}^n \frac1k$.  Inasmuch as $H_n=\log(n)+\gamma +O(1/n)$, the series converges.  Therefore we can write
$$\begin{align}
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}H_n&=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\sum_{k=1}^n \frac1k\\\\
&=\sum_{k=1}^\infty \frac1k \sum_{n=k}^\infty \frac{(-1)^{n+1}}{n+1}\\\\
&=\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{(-1)^{n+k}}{k(n+k)}\tag1\\\\
&=\sum_{k=1}^\infty \sum_{n=1}^\infty \left(\frac{(-1)^{n+k}}{nk}-\frac{(-1)^{n+k}}{n(n+k)}\right)\tag2 \\\\
&=\log^2(2)-\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{(-1)^{n+k}}{n(n+k)}\tag3
\end{align}$$



The interchange of the order of summation is straightforward to justify and is left to the reader to verify.  In going from $(1)$ to $(2)$ we used partial fraction expansion.



Now, not that the series on the right-hand side of $(3)$ is identical to the series on the right-hand side of $(1)$ (indices interchanged).  Hence, we see that
$$2\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{(-1)^{n+k}}{k(n+k)}=\log^2(2)$$
from which we find that
$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}H_n=\frac12\log^2(2)}$$
as was to be shown!  And we are done.
