Exact Value of a Series

Question

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.

Answer 1

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

Answer 2

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.