calculus - Are there any known methods to compute series $sum_{k=0}^{infty}2^k big(sum_{n=0}^{infty}frac{(-1)^n}{(n2^{k+1}+(2k+1))^3}big)$?

I would like to ask if there are any known methods to compute series like this one ?

$$\sum_{k=0}^{\infty}2^k \bigg(\sum_{n=0}^{\infty}\frac{(-1)^n}{(n2^{k+1}+(2k+1))^3}\bigg)$$
And their names so i can look for them if they exist.
I never studied double sums before that's why i am asking, thanks in advance.

Answer

We have

$\displaystyle \sum\limits_{k=0}^\infty 2^k \sum\limits_{n=0}^\infty \frac{(-1)^n}{(n2^{k+1}+2k+1)^3} =\sum\limits_{k=0}^\infty \frac{2^k}{(2k+1)^3} - \sum\limits_{k=0}^\infty \left(\frac{1}{2^{2k+3}}\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{(n+\frac{2k+1}{2^{k+1}})^3}\right)$

and

$\displaystyle\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{(n+a)^3}=\frac{1}{8}\left(\zeta(3,\frac{1}{2}+\frac{a}{2})-\zeta(3,1+\frac{a}{2})\right)\,$ . $\hspace{1cm}$ (see: Hurwitz zeta function)

It follows

$\displaystyle\sum\limits_{k=0}^\infty \frac{1}{2^{2k+3}}\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{(n+\frac{2k+1}{2^{k+1}})^3}=\sum\limits_{k=0}^\infty \frac{1}{2^{2k+6}}\left(\zeta(3,\frac{1}{2}+\frac{2k+1}{2^{k+2}})-\zeta(3,1+\frac{2k+1}{2^{k+2}})\right)$

$\hspace{5.5cm}\approx 0.05957499...$

$\displaystyle\sum\limits_{k=0}^\infty \frac{2^k}{(2k+1)^3}$ is divergent because $\displaystyle\left(\frac{2^k}{(2k+1)^3}\right)_k$ is not a null sequence and

$\displaystyle\sum\limits_{k=0}^\infty \left(\frac{1}{2^{2k+3}}\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{(n+\frac{2k+1}{2^{k+1}})^3}\right)$ is convergent therefore the initial series is divergent.