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?
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.)
