Proving that $\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^n}{(k+1)\times (n-k+1)}$ converges How can I prove that the following summation converges?
$$\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^n}{(k+1)\times (n-k+1)}$$
I tried to prove that by proving that the following summation in in absolute value converges so the original one converges too, but that's incorrect.
Any other ideas?
 A: If we write it this way
$$\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^n}{(k+1)\times (n-k+1)} = \sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^k}{(k+1)}\times \frac{(-1)^{n-k}}{(n-k+1)}$$
Then this is the Cauchy product (discrete convolution) of $\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)}$ with itself. We know that $\ln(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k+1}x^k}{k}$, by changing the index $k \rightarrow k+1$ we get $\ln(1+x)=\sum_{k=0}^\infty \frac{(-1)^{k}x^{k+1}}{k+1}$ and we have $\ln(2)=\ln(1+1)=\sum_{k=0}^\infty \frac{(-1)^{k}}{k+1}$ and we can conclude that 
$$\sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^n}{(k+1)\times (n-k+1)} = \left(\ln(2)\right)^2$$
For the convergence part, this series $\color{red}{\text{does not}}$ converge absolutely. To prove this, we consider the tail's sum
$$ \sum_{r=m}^n \sum_{k=0}^r \Bigl|\frac{(-1)^r}{(k+1) (r-k+1)} \Bigr| = \sum_{r=m}^n\sum_{k=0}^r  \frac{1}{(k+1) (r-k+1)} $$
Now consider the inner sum 
$$\sum_{k=0}^r \frac{1}{(k+1) (r-k+1)}  = \frac{1}{r+2}\sum_{k=0}^r \frac{(r-k+1) + (k+1)}{(k+1) (r-k+1)} \\ = \frac{1}{r+2}\sum_{k=0}^r\Bigl( \frac{1}{k+1}+\frac{1}{r-k+1} \Bigr) = \frac{2}{r+2}\sum_{k=0}^r \frac{1}{k+1} > \frac{2\ln(r+2)}{r+2}$$
using the integral test, because $ \sum_{k=0}^r \frac{1}{k+1} = \sum_{k=1}^{r+1} \frac{1}{k} > \int_1^{r+2} \frac{1}{x} dx = \ln(r+2)$
and overall we have 
$$\sum_{r=m}^n\sum_{k=0}^r  \frac{1}{(k+1) (r-k+1)} > 2\sum_{r=m}^n \frac{\ln(r+2)}{r+2}$$
which we know diverges when $n\rightarrow \infty$ regardless to $m$ (Rudin, Principles of Mathematical Analysis, Theorem 3.29. for $p=-1$). But this series is alternating and it converges non-absolutely (again, Rudin, Principles of Mathematical Analysis, Theorem 3.43.). To show this, the only thing we need to show is $\sum_{k=0}^r \frac{1}{(k+1) (r-k+1)} \ge \sum_{k=0}^{r+1} \frac{1}{(k+1) (r-k+2)}$ because the alternating series is of the form of $\sum_{n=0}^\infty (-1)^n \Bigl( \sum_{k=0}^n \frac{1}{(k+1)\times (n-k+1)}\Bigr)$. Based on the above calculation we have to show
$$\frac{2}{r+2}\sum_{k=0}^r \frac{1}{k+1} \ge \frac{2}{r+3}\sum_{k=0}^{r+1} \frac{1}{k+1} \\ \Rightarrow \frac{1}{r+2}\sum_{k=0}^r \frac{1}{k+1} \ge \frac{1}{r+3}\sum_{k=0}^{r} \frac{1}{k+1} + \frac{1}{(r+2)(r+3)}  \\ \Rightarrow \Bigl( \sum_{k=0}^{r} \frac{1}{k+1}  \Bigr)\Bigl(\frac{1}{r+2} - \frac{1}{r+3}\Bigr)\ge \frac{1}{(r+2)(r+3)} \\ \Rightarrow \Bigl( \sum_{k=0}^{r} \frac{1}{k+1}  \Bigr)\Bigl(\frac{1}{(r+2)(r+3)}\Bigr)\ge \frac{1}{(r+2)(r+3)} $$
But that's obviously true $\square$. 
Therefore the value of the series can be anything according to the Riemann's rearrangement theorem, but in this formation, the value of the series converges to $\left(\ln(2)\right)^2$
(P.S. if you want to get an intuition, think of it as $\int\int \frac{1}{x^2}$ that gives you $\ln x$ which we know is unbounded.)
