Evaluate:: $ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 +\cdots + \frac 1n\right) $ How to evaluate the series:
$$  2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 + \cdots + \frac 1n\right) $$
According to Mathematica, this converges to $ (\log 2)^2 $.
 A: In this answer, it is shown that
$$
\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}H_n=\frac12\zeta(2)-\frac12\log(2)^2
$$
This sum is
$$
\begin{align}
2\sum_{n=1}^\infty\frac{(-1)^n}{n}H_{n-1}
&=2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\left(\frac1n-H_n\right)\\
&=2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}-2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}H_n\\
&=\zeta(2)-\left(\zeta(2)-\log(2)^2\right)\\[6pt]
&=\log(2)^2
\end{align}
$$
A: Recall that, formally,
$$
\left(\sum_{n=1}^{\infty} a_n\right)\left(\sum_{n=1}^{\infty} b_n\right) = \sum_{n=1}^{\infty} c_{n+1},$$
where
$$
c_n = \sum_{k=1}^{n-1} a_k b_{n-k}.
$$
If the series $\sum c_{n+1}$ converges, then the above equality is actually true.  You seem to know how to show this, so I'll just demonstrate the formal aspect of the problem.
Let $a_n = b_n = \frac{(-1)^{n}}{n}$.  Then
$$
a_k b_{n-k} = \frac{(-1)^n}{k(n-k)} = \frac{(-1)^n}{n}\left(\frac{1}{k}+\frac{1}{n-k}\right),
$$
so that
$$
\begin{align*}
c_n &= \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \left(\frac{1}{k}+\frac{1}{n-k}\right) \\
&= 2\frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{1}{k}.
\end{align*}
$$
We therefore have
$$
2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1} \sum_{k=1}^{n} \frac{1}{k} = \left(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)^2 = (-\log 2)^2 = (\log 2)^2.
$$
A: Here is another approach that I just noticed
$$
\begin{align}
2\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n+1}\sum_{k=1}^n\frac1k
&=2\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{(-1)^{n+1}}{k(n+1)}\tag{1}\\
&=2\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{n+k}}{k(n+k)}\tag{2}\\
&=2\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{n+k}\frac1n\left(\frac1k-\frac1{n+k}\right)\tag{3}\\
&=\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{n+k}\frac1n\frac1k\tag{4}\\[6pt]
&=\log(2)^2\tag{5}
\end{align}
$$
Explanation:
$(1)$: change the order of summation
$(2)$: substitute $n\mapsto n+k-1$
$(3)$: partial fractions
$(4)$: swap $n$ and $k$ in $(2)$ add to $(3)$ and divide by $2$
$(5)$: $\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}}{n}=\log(2)$
A: Use generating functions:
Consider
$$-\log(1-x) = \sum_{n=1}^\infty \frac{x^n}{n}.$$
Dividing by $1-x$, we get
$$-\frac{\log(1-x)}{1-x} = \sum_{n=1}^\infty \left(\sum_{k=1}^n \frac{1}{k}\right)x^n.$$
Integrating this and multiplying everything by $2$ gives
$$\left[\log(1-x)\right]^2 = 2\sum_{n=1}^\infty \left(\sum_{k=1}^n \frac{1}{k}\right)\frac{x^{n+1}}{n+1} + C,$$
where $C$ is some constant. But we can get rid of $C$ by plugging $x=0$ into both sides, which gives $C=0$:
$$\left[\log(1-x)\right]^2 = 2\sum_{n=1}^\infty \left(\sum_{k=1}^n \frac{1}{k}\right)\frac{x^{n+1}}{n+1}.$$
From here, we'd like to simply plug in $x=-1$ and say our answer is $(\log{2})^2$, but we have to first check to make sure the power series on the right actually converges there. To do this, set $H_n=1+\frac{1}{2}+\cdots + \frac{1}{n}$ (the "$H$" is for "harmonic", since $H_n$ is the $n$th harmonic number). Let's see when the inequality
$$ \frac{(n+1)H_{n+1}}{(n+2)H_n}<1$$
holds. Rearranging terms, and using the fact that $H_{n+1}=H_n+\frac{1}{n+1}$, it follows that the above inequality holds exactly when $H_n>1$. But a quick glance at the definition of $H_n$ shows that this is always true! Therefore, the terms of our series decrease in absolute value. Since they also converge to zero (they're all less than $1/(n+1)$, which converges to zero), the entire series converges by the alternating series test.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&2\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n + 1}
\pars{1 + {1 \over 2} + \cdots + {1 \over n}} =
-2\sum_{n = 1}^{\infty}\pars{-1}^{n}\, H_{n}\int_{0}^{1}x^{n}\,\dd x =
-2\int_{0}^{1}\sum_{n = 1}^{\infty}H_{n}\pars{-x}^{n}\,\dd x
\\[5mm] = &\
-2\int_{0}^{1}\braces{-\,{\ln\pars{1 - \bracks{-x}} \over 1 - \pars{-x}}}\,\dd x =
2\int_{0}^{1}{\ln\pars{1 +x} \over 1 + x}\,\dd x =
\left.\ln^{2}\pars{1 + x}\,\right\vert_{\ x\ =\ 0}^{\ x\ =\ 1} =
\bbx{\ds{\ln^{2}\pars{2}}}
\end{align}
A: This is a special case of a more general result derived here.
$$S = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n+1} \sum_{k=1}^n \dfrac1k$$
Recall that $\dfrac1k = \displaystyle \int_0^1 x^{k-1} dx$ and $\dfrac1{n+1} = \displaystyle \int_0^1 y^n dy$.
Now use the following fact.
$$\sum_{k=0}^{\infty} \int_0^1 (-z)^k dz = \lim_{n \to \infty} \int_0^1 \dfrac{1 - (-z)^n}{1+z} dz$$
The sequence of  functions $f_n(z) = \dfrac{1 - (-z)^n}{1+z}$ is dominated by the function $g(z) = \dfrac2{1+z}$ in the interval $[0,1]$, which is integrable. Hence, we can swap the limit and the integral to get that
$$\lim_{n \to \infty} \int_0^1 \dfrac{1 - (-z)^n}{1+z} dz = \int_0^1 \dfrac{dz}{1+z}$$
Hence,
$$S = \sum_{n=1}^{\infty} (-1)^{n+1} \int_0^1 y^n dy \left(\sum_{k=1}^n \int_0^1 x^{k-1} dx  \right) = \sum_{n=1}^{\infty} (-1)^{n+1} \int_0^1 y^n dy \left(\int_0^1 \dfrac{1-x^n}{1-x} dx  \right)$$
Hence,
$$S = \int_0^1 \int_0^1 \dfrac{\dfrac{y}{1+y} - \dfrac{xy}{1+xy}}{1-x} dy dx = \int_0^1 \int_0^1 \dfrac{y+xy^2-xy-xy^2}{(1+y)(1+xy)(1-x)} dx dy\\
=\int_0^1 \int_0^1 \dfrac{y}{(1+y)(1+xy)} dx dy = \int_0^1 \dfrac{\log(1+y)}{1+y} dy = \left. \dfrac{\log^2(1+y)}2 \right \vert_0^1 = \dfrac{\log^2(2)}2$$
The sum you are interested in is $2S$ and hence the answer is $\log^2(2)$.
