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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 !

Klangen
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Hptunjy Prjkeizg
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  • I took the risk to add some tags ... so the series converges absolutely and rearranging the terms is fine https://en.wikipedia.org/wiki/Absolute_convergence ... until $(1)$. However $(2)$ seems to contain a typo, can you elaborate it? – rtybase Sep 14 '17 at 20:12
  • No, $(2)$ seems to be fine too ... digging further ... – rtybase Sep 14 '17 at 20:16
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    the sum $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^3}$$ is = to the alternate sum $$\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n-1)^3}$$ + 2 times all the negative terms. – Hptunjy Prjkeizg Sep 14 '17 at 20:18
  • https://en.wikipedia.org/wiki/Van_Wijngaarden_transformation – jimjim Sep 14 '17 at 21:52
  • I think you last two series are wrong as stated as they leave out the final non-alternating term included in equation (2). Without this term your sum diverges thus for example for any $v$. $(\zeta(3))=\sum_{k=0}^{v-1} 2^k \sum_{n=1}^{\infty} \frac{(-1)^n}{(2^{k+1}n-(2^{k+1}-1))^3}+2^v \frac{1}{(2^{v+1}n-(2^{v+1}-1))^3}$. Your series cannot be converted to an infinite series in the way you describe as you always need the non-alternating final balancing term. – James Arathoon Sep 18 '17 at 09:37
  • sorry I meant $\frac{7}{8}\zeta(3)$ in above comment. – James Arathoon Sep 18 '17 at 09:52
  • i think the second term doesn't exist if it's an infinite expression ? – Hptunjy Prjkeizg Sep 18 '17 at 17:57
  • That would be the case if the first term converged to $\frac{7}{8}\zeta(3) $ in the limit $v$ tends to infinity, but that does not happen. For example for $v=20$ the first term is approximately -1048574.9482002 and the second term 1048576 $(2^{20})$ giving the approximate value for $\frac{7}{8}\zeta(3)$ of 1.0517998. The second term simplifying to $2^v$, in the limit v becomes large giving $(\frac{7}{8}\zeta(3))=\sum_{k=0}^{v-1} 2^k \sum_{n=1}^{\infty} \frac{(-1)^n}{(2^{k+1}n-(2^{k+1}-1))^3}+2^v$ – James Arathoon Sep 18 '17 at 20:46

1 Answers1

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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.

James Arathoon
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