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 $.

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

$$ 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1} \sum_{k=1}^{n} \frac{1}{k} = \left(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)^2 = (-\log 2)^2 = (\log 2)^2. $$

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.

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 $\log^2(2)$.

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 $$ \begin{align} 2\sum_{n=1}^\infty\frac{(-1)^n}{n}H_{n-1} &=2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\left(\frac1n-H_n\right)\\ &=2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}-2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}H_n\\ &=\zeta(2)-\left(\zeta(2)-\log(2)^2\right)\\[6pt] &=\log(2)^2 \end{align} $$

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 $2$
$(5)$: $\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}}{n}=\log(2)$