Use integration to estimate $\sum\limits_{n=0}^{+\infty}(-1)^n\frac 1 {n+1}\sum\limits_{k=0}^n \frac 1 {k+1}$

Recall the geometric series:

$$\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}$$

Integrate both sides from zero to one,

$$\sum_{k=0}^n\frac1{k+1}=\int_0^1\frac{1-x^{n+1}}{1-x}\ dx$$

It thus follows that

$$v_n=\int_0^1\frac{1-x^{n+1}}{(n+1)(1-x)}\ dx$$

Apply integration by parts to get

$$v_n=-\int_0^1x^n\ln(1-x)\ dx$$

for every $n$, hence $$S=-\int_0^1\frac{\ln(1-x)}{1+x}\ dx$$

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.