Show that $$\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^s}=\frac{2^{s-1}-1}{2^{s-1}}\sum\limits_{n=1}^{\infty}\dfrac{1}{n^s}$$

for all integers $s1$

Question

Show that $$\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^s}=\frac{2^{s-1}-1}{2^{s-1}}\sum\limits_{n=1}^{\infty}\dfrac{1}{n^s}$$

for all integers $s>1$

myininaya · Answer

eww do we have to use the reimaann zeta function thingy ?

myininaya · Answer

and however it is spelled

myininaya · Answer

hmmm maybe we can use induction

myininaya · Answer

that still might require that zeta function thing

ganeshie8 · Answer

Oh I think induction should work xD but i havent tried induction yet... 

it is a relation between riemann zeta and dirichlet eta functions
but that relation can be derived using algebra w/o mentioning about zeta..

ganeshie8 · Answer

came across this nice relation while working on @xapproachesinfinity 's latest problem..

zarkon · Answer

expand the right hand side...works out pretty easily

thadyoung · Answer

@bibby

ganeshie8 · Answer

Wow yes that should be the simplest way to work this
$$\begin{align}\frac{2^{s-1}-1}{2^{s-1}}\sum\limits_{n=1}^{\infty}\dfrac{1}{n^s}&=\left(1-\frac{1}{2^{s-1}}\right)\sum\limits_{n=1}^{\infty}\dfrac{1}{n^s}\~\
&=\sum\limits_{n=1}^{\infty}\dfrac{1}{n^s}-\frac{1}{2^{s-1}}\sum\limits_{n=1}^{\infty}\dfrac{1}{n^s}\~\

&=\sum\limits_{n=1}^{\infty}\dfrac{1}{n^s}-\sum\limits_{n=1}^{\infty}\dfrac{2}{(2n)^s}\~\
&=\left(\sum\limits_{n=1}^{\infty}\dfrac{1}{n^s}-\sum\limits_{n=1}^{\infty}\dfrac{1}{(2n)^s}\right) - \sum\limits_{n=1}^{\infty}\dfrac{1}{(2n)^s} \~\
&=\left(\sum\limits_{n=1}^{\infty}\dfrac{1}{(2n-1)^s}\right) - \sum\limits_{n=1}^{\infty}\dfrac{1}{(2n)^s} \~\
&=\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^s} \~\
\end{align}$$