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.
 A: 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.
A: I wanted to get an other form for the $\zeta (3)$.
$$\sum_{n=1}^{\infty}\frac{1}{n^3}\space =\space \sum_{n=0}^{\infty}\frac{1}{(2n+1)^3}\space + \sum_{n=1}^{\infty}\frac{1}{(2n)^3}$$
Or:
$$\sum_{n=1}^{\infty}\frac{1}{n^3}\space - \frac{1}{8}\sum_{n=1}^{\infty}\frac{1}{n^3} \space = \sum_{n=0}^{\infty}\frac{1}{(2n+1)^3}$$
$$\frac{7}{8}\sum_{n=1}^{\infty}\frac{1}{n^3}\space = \sum_{n=0}^{\infty}\frac{1}{(2n+1)^3} \tag{1}$$
$$\frac{7}{8}\sum_{n=1}^{\infty}\frac{1}{n^3}\space = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^3}\space + \space 2\sum_{n=0}^{\infty}\frac{1}{(4n+3)^3} \tag{2}$$
What i'm trying to do is to express it using a sum of alternate sums. 
$$\frac{7}{8}\sum_{n=1}^{\infty}\frac{1}{n^3}\space = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^3}\space + \space 2\sum_{n=0}^{\infty}\frac{(-1)^n}{(4n+3)^3}\space +\space 4\sum_{n=0}^{\infty}\frac{(-1)^n}{(8n+7)^3}\space + \space ... \space +\space 2^k\sum_{n=0}^{\infty}\frac{(-1)^n}{(2^{k+1}n+(2^{k+1}-1))^3}$$
Which is a double sum : 
$$\frac{7}{8}\sum_{n=1}^{\infty}\frac{1}{n^3}\space = \space \sum_{k=0}^{\infty}2^k\sum_{n=0}^{\infty}\frac{(-1)^n}{(2^{k+1}n+(2^{k+1}-1))^3}$$
that is actually what $\zeta(3)$ is equal to, i typed it wrong in the question <. < and maybe that is why it's divegent !
