How can I show that these sum are equivalent and what is the closed form of $F(x)$?
$$\sum_{j=2}^{\infty}{1\over j^x(j^2+1)}=\sum_{i=1}^{\infty}(-1)^{i-1}\zeta(2i+x)=F(x)$$
$x\ge0$
My try:
Setting $x=0$ [Just to do an example to see where I can get to...]
$$\sum_{j=2}^{\infty}{1\over j^2+1}=\sum_{i=1}^{\infty}(-1)^{i-1}\zeta(2i)$$
We know that $$\sum_{n=1}^{\infty}[\zeta(2n)-1]={3\over 4}$$
Add them and we have, I can't see any link at all!
$$\sum_{i=1}^{\infty}(-1)^{i-1}\zeta(2i)+\sum_{n=1}^{\infty}[\zeta(2n)-1]=2\sum_{n=1}^{\infty}[\zeta(4n-2)-1]$$
I found this formula in wolfram
$$\sum_{k=1}^{\infty}{1\over k^2-x^2}=\sum_{n=0}^{\infty}\zeta(2n+2)x^{2n}={1-x\pi\cot{x\pi}\over 2x^2}\tag1$$
Setting $x=i=\sqrt{-1}$
We got
$${1\over 2}+\sum_{k=2}^{\infty}{1\over k^2+1}=\sum_{n=0}^{\infty}(-1)^n\zeta(2n+2)={1-i\pi\cot{i\pi}\over -2}$$
So we have
$$\sum_{k=2}^{\infty}{1\over k^2+1}=\sum_{n=0}^{\infty}(-1)^n\zeta(2n+2)={-2+i\pi\cot{i\pi}\over 2}$$
How do we used this formula (1) to find a closed form for $F(x)$?