Exact Value of a Series It is very easy to show that the series 
$$\frac{1-1/2}{1\times2} - \frac{1-1/2+1/3}{2\times3} + \frac{1-1/2+1/3-1/4}{3\times4} - ...$$
i.e.
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}[1-\frac{1}{2} + \frac{1}{3} - ...+ \frac {(-1)^{n}}{n+1}]$$
is convergent. Can one find its exact value? Or is it unreasonable to hope for such a thing?
Thank you for your answers.
 A: The interior sum is equal to $\int_0^1\frac{1-(-x)^{n+1}}{1+x}dx$ and 
$$
\sum _{n=1}^{\infty } \frac{(-1)^{n+1} \left(1-(-x)^{n+1}\right)}{n (n+1) (x+1)}=
\frac{(x-1) \log (1-x)-x+\log (4)-1}{1+x}.
$$
So the answer is eqaul to
$$
\int_0^1 \frac{(x-1) \log (1-x)-x+\log (4)-1}{1+x}dx=\frac{\pi ^2}{6}+\log ^2(2)-2.
$$
A: Since
$$\frac1{n(n+1)}=\frac1n-\frac1{n+1},$$
we have
$$\begin{aligned}
\sum_{n=1}^N\sum_{k=1}^{n+1}\frac{(-1)^{n+k}}{kn(n+1)}
&=\sum_{n=1}^N\sum_{k=1}^{n+1}\frac{(-1)^{n+k}}{kn}+\sum_{n=2}^{N+1}\sum_{k=1}^n\frac{(-1)^{n+k}}{kn}\\\\
&=2\sum_{n=1}^N\sum_{k=1}^n\frac{(-1)^{n+k}}{kn}-\sum_{n=1}^N\frac1{n(n+1)}-1+\sum_{k=1}^{N+1}\frac{(-1)^{N+1+k}}{k(N+1)}\\\\
&=\sum_{n,k=1}^N\frac{(-1)^{n+k}}{kn}+\sum_{n=1}^N\frac1{n^2}-2+\frac1{N+1}+\frac{(-1)^{N+1}}{N+1}\sum_{k=1}^{N+1}\frac{(-1)^k}k\\\\
&=\left(\sum_{n=1}^N\frac{(-1)^{n}}n\right)^2+\sum_{n=1}^N\frac1{n^2}-2+\frac1{N+1}+\frac{(-1)^{N+1}}{N+1}\sum_{k=1}^{N+1}\frac{(-1)^k}k.
\end{aligned}$$
As $\sum_{n=1}^\infty\frac{(-1)^n}n=-\log2$ and $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$, this implies
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}\sum_{k=1}^{n+1}\frac{(-1)^{k+1}}k=(\log2)^2+\frac{\pi^2}6-2.$$
There may be a numerical error somewhere, but in principle the method should work.
