# Thread: zeta function series

1. ## zeta function series

I want to show that $\displaystyle \displaystyle \sum^{\infty}_{n=1} \big( \zeta(4n)-1 \big) = \frac{7- 2 \coth \pi}{8}$

$\displaystyle = \displaystyle \sum^{\infty}_{n=1} \sum^{\infty}_{m=2} \frac{1}{m^{4n}}$

Because I'm summing over only positive values, I think I can switch the order of summation.

$\displaystyle \displaystyle = \sum^{\infty}_{m=2} \sum^{\infty}_{n=1} \frac{1}{m^{4n}} = \sum^{\infty}_{m=2} \sum^{\infty}_{n=1} \Big(\frac{1}{m^{4}}\Big)^{n}$

$\displaystyle \displaystyle = \sum^{\infty}_{m=2} \frac{\frac{1}{m^{4}}}{1- \frac{1}{m^{4}}} = \sum^{\infty}_{m=2} \frac{1}{m^{4}-1}$

$\displaystyle = \displaystyle \sum_{m=2}^{\infty} \Big( \frac{1}{4(m-1)} - \frac{1}{4(m+1)} - \frac{1}{2(m^{2}+1)} \Big)$

$\displaystyle = \displaystyle \frac{1}{4} \sum_{m=2}^{\infty} \Big( \frac{1}{(m-1)} - \frac{1}{(m+1)} \Big) - \frac{1}{2} \sum^{\infty}_{m=2} \frac{1}{(m^{2}+1)}$

which I think is OK since both series converge

The first series is telescoping and evaluates to $\displaystyle \displaystyle \frac{3}{2}$ .

The second series is where I run into problems.

EDIT: $\displaystyle \displaystyle \sum^{\infty}_{m=0} \frac{1}{(m^{2}+1)} = \sum^{\infty}_{m=0} \frac{1}{(m^{2}+1)} - \frac{3}{2}$

$\displaystyle \displaystyle = \frac{1}{2i} \sum^{\infty}_{m=0} \Big(\frac{1}{m-i} - \frac{1}{m+i} \Big) - \frac{3}{2}$

At this point I probably need to use the digamma function, but I really don't know.

2. You can apply Parseval to the function $\displaystyle t\mapsto e^{-t}$.

3. Originally Posted by girdav
You can apply Parseval to the function $\displaystyle t\mapsto e^{-t}$.
Parseval's identity?

4. Yes.