I wanted to get an other form for $\zeta (3)$, where $\zeta$ denotes the Riemann Zeta function: $$\sum_{n=1}^{\infty}\frac{1}{n^3}\space =\space \sum_{n=1}^{\infty}\frac{1}{(2n-1)^3}\space + \sum_{n=1}^{\infty}\frac{1}{(2n)^3},$$ or, $$\sum_{n=1}^{\infty}\frac{1}{n^3}\space - \frac{1}{8}\sum_{n=1}^{\infty}\frac{1}{n^3} \space = \sum_{n=1}^{\infty}\frac{1}{(2n-1)^3}$$ $$\frac{7}{8}\sum_{n=1}^{\infty}\frac{1}{n^3}\space = \sum_{n=1}^{\infty}\frac{1}{(2n-1)^3} \tag{1}$$ $$\frac{7}{8}\sum_{n=1}^{\infty}\frac{1}{n^3}\space = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n-1)^3}\space + \space 2\sum_{n=1}^{\infty}\frac{1}{(4n-3)^3} \tag{2}$$ What i'm trying to do is to express it using only alternate sums.
$$\frac{7}{8}\sum_{n=1}^{\infty}\frac{1}{n^3}\space = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n-1)^3}\space + \space 2\sum_{n=1}^{\infty}\frac{(-1)^n}{(4n-3)^3}\space +\space 4\sum_{n=1}^{\infty}\frac{(-1)^n}{(8n-7)^3}\space + \space ... \space +\space 2^k\sum_{n=1}^{\infty}\frac{(-1)^n}{(2^{k+1}n-(2^{k+1}-1))^3}$$ Which is a double sum : $$\frac{7}{8}\sum_{n=1}^{\infty}\frac{1}{n^3}\space = \space \sum_{k=0}^{\infty}2^k\sum_{n=1}^{\infty}\frac{(-1)^n}{(2^{k+1}n-(2^{k+1}-1))^3}$$ I was hoping to get somthing simpler than the first sum, even though i know that alternate sums are easyer to compute, that is why i wanted to express it using them, but is this right ? did i make any mistake in any step ? thanks in advance !
The maths is clearer if you work this out more generally.
Lets define $\zeta(k)=\sum_{n=1}^{\infty}\frac{1}{n^k}$ for $k\ge2$ and we want to calculate both $\lambda(k)=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^k}$ and $\eta(k)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^k}$ in terms of $\zeta(k)$.
Start with finding $\zeta(k)-\lambda(k)$, that is
$$\zeta(k)-\lambda(k)=\frac{1}{2^k}\left(\frac{1}{1^k}+\frac{1}{2^k}+\frac{1}{3^k}+...\right)=\frac{1}{2^k}\zeta(k)$$
Rearranging this gives $$\lambda(k)=\frac{2^k-1}{2^k}\zeta(k)\tag1$$
In the case of $k=3$ this is $$\lambda(3)=\frac{7}{8}\zeta(3)$$
The $\eta(k)$ is found using $\eta(k)=2\lambda(k)-\zeta(k)$ which if we substitute in (1) immediately leads to
$$\eta(k)=\left( 2\frac{2^k-1}{2^k}-1 \right)\zeta(k)$$
Simplifying to
$$\eta(k)=\left( 1-2^{1-k} \right)\zeta(k)\tag2$$
In the case of $k=3$ this is $$\eta(3)=\frac{3}{4}\zeta(3)$$
This last formula gives the simplest sum for alternating terms.