TITLE: Use integration to estimate $\sum\limits_{n=0}^{+\infty}(-1)^n\frac 1 {n+1}\sum\limits_{k=0}^n \frac 1 {k+1}$
QUESTION [8 upvotes]: Let $$v_n=\dfrac 1 {n+1}\sum_{k=0}^n \dfrac 1 {k+1}$$
We wanna study the sum $$S=\sum_{n=0}^{+\infty}(-1)^n v_n$$
The problem says we should first find $\omega(x)$ s.t. 
$$v_n=\int_0^1 x^n\omega(x)dx$$
Then we'll have $S=\int_0^1\dfrac {\omega(x)} {1+x}dx$, but I can't find such $\omega(x)$. What's the idea of constructing such integral?

REPLY [5 votes]: After we get
$$ S = -\int_{0}^{1}\frac{\log(1-x)}{1+x}\,dx = -\int_{0}^{1}\frac{\log x}{2-x}\,dx =-\left.\frac{d}{d\alpha}\int_{0}^{1}\frac{x^{\alpha}\,dx}{2-x}\,\right|_{\alpha=0^+}$$
we also have
$$ S = \text{Li}_2\left(\frac{1}{2}\right) = \color{red}{\frac{\pi^2}{12}-\frac{\log(2)^2}{2}}$$
by the dilogarithm reflection formula.