2

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 $F(x)$?

Did
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    The series $\sum_{k=1}^{\infty} (-1)^{k-1} \zeta(x+2k)$ does not converge in ordinary sense. I can check that the LHS equals $\sum_{k=1}^{\infty} (-1)^{k-1} [\zeta(x+2k) - 1]$, so I guess you made a mistake when transcribing the problem. – Sangchul Lee Jan 01 '17 at 10:23
  • For the first part, use $$\frac1{j^x(j^2+1)}=\frac1{j^{x+2}(1+\frac1{j^2})}=\frac1{j^{x+2}}\sum_{i=0}^{\infty}(-1)^i\frac1{j^{2i}}=\sum_{i=1}^{\infty}(-1)^{i-1}\frac1{j^{x+2i}}$$ for every $j\geqslant2$. – Did Jan 01 '17 at 10:31
  • i am still curious, where do you find all this identities? Namagiri? – tired Jan 01 '17 at 10:34
  • Take a look at Jack D'Aurizio's answer here http://math.stackexchange.com/questions/1418568/closed-form-for-sum-k-1-infty-zeta4k1-1?rq=1 – Marco Cantarini Jan 01 '17 at 11:32
  • @X-Men i see, a real enthusiast. keep going, your posts are an enrichment to this platform! – tired Jan 01 '17 at 11:52

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