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.

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
$$
\int_0^1 \frac{(x-1) \log (1-x)-x+\log (4)-1}{1+x}dx=\frac{\pi ^2}{6}+\log ^2(2)-2.
$$
