Does $\sum _{n=1}^{\infty }\sum _{i=1}^{n }\frac{\left(-1\right)^n}{i\cdot n}$ have a finite value? If so, evaluate its closed form.
I'm pretty sure its related to $\begin{array}{l}\zeta \left(2\right)=\frac{\pi ^2}{6}\end{array}$
But i am having trouble converting this into a form that can be further manipulated.
Thanks ☺☺☺
Consider $$\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty\right)^2=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}...\infty+2\left(1\left(-\frac{1}{2}\right)+1\left(\frac{1}{3}\right)+...\ +-\frac{1}{2}\left(\frac{1}{3}\right)+-\frac{1}{2}\left(-\frac{1}{4}\right)...\infty\right)$$
The numbers in the brackets are the sum of products of unique pairs of all numbers.*(-1)^(sum of both numbers)
We can rearrange it so that the denominators are in a sequence.
Let $$x=-1\left(-\frac{1}{2}\right)+1\left(\frac{1}{3}\right)+...\ +-\frac{1}{2}\left(\frac{1}{3}\right)+-\frac{1}{2}\left(-\frac{1}{4}\right)...\infty $$ $$=\left(-\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}...\infty \right)+\left(\left(\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 8}...\infty \right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...\right)\right)-\left(\frac{1}{1\cdot 4}+...and\ so\ on\right)$$
Now each term telescopes, $$T_n=\frac{1}{n}\sum _{i=1}^n\frac{1}{i}$$ This is equivalent to the nth term in the question, So the sum is $$\begin{array}{l}\frac{\left(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty \right)^2-\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}...\infty \right)\right)}{2}=\frac{\left(\ln \left(2\right)\right)^2}{2}-\frac{\pi ^2}{12}\end{array}$$