It is very easy to show that the series $$\frac{11/2}{1\times2}  \frac{11/2+1/3}{2\times3} + \frac{11/2+1/31/4}{3\times4}  ...$$ 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 $$ \sum _{n=1}^{\infty } \frac{(1)^{n+1} \left(1(x)^{n+1}\right)}{n (n+1) (x+1)}= \frac{(x1) \log (1x)x+\log (4)1}{1+x}. $$ So the answer is eqaul to $$ \int_0^1 \frac{(x1) \log (1x)x+\log (4)1}{1+x}dx=\frac{\pi ^2}{6}+\log ^2(2)2. $$ 


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 $$\sum_{n=1}^\infty\frac{(1)^{n+1}}{n(n+1)}\sum_{k=1}^{n+1}\frac{(1)^{k+1}}k=(\log2)^2+\frac{\pi^2}62.$$ There may be a numerical error somewhere, but in principle the method should work. 

