Evaluate the sum 1/(k^2+1) from k= 1 to infinity?

Question

Find the sum 1/(k^2+1) from k = 1 to infinity 
Give sum 1/k^2 = pi^2/6, sum 1/k^4 = pi^4/90, sum 1/k^6 = pi^6/945 all from k = 1 to infinity 

i know sum 1/(k^2+1) = sum (1/k^2 - 1/(k^2*(k^2+1)))    = sum 1/k^2 - sum 1/(k^2(k^2+1))
= pi^2/6 -sum (1/(k^4+k^2))
i dont know what to do from here, 
can anybody help me on this, Thanks

kb · Accepted Answer

One way to evaluate this cleanly is to use Complex Analysis.

Here's the key result (proved using the Residue Theorem):
Under "reasonable" growth conditions on f, we have
Σ(-∞ to ∞) f(k) = - {sum of the residues of π cot(πz) f(z) at the poles of f(z)}.
------------------------------
Here, f(z) = 1/(z^2 + 1), which has simple poles at z = ± i.

The residue of π cot(πz) * 1/(z^2 + 1) at z = i equals
lim(z→i) (z - i) * π cot(πz)/(z^2 + 1) 
= lim(z→i) π cot(πz)/(z + i) 
= π cot(πi)/(2i) 
= (-π/2) coth π.

Similarly, the residue of π cot(πz) * 1/(z^2 + 1) at z = i equals (-π/2) coth π as well.

Hence,
Σ(-∞ to ∞) 1/(k^2 + 1) = - {(-π/2) coth π + (-π/2) coth π} = π coth π.

However,
Σ(-∞ to ∞) 1/(k^2 + 1) = Σ(-∞ to -1) 1/(k^2 + 1) + 1/(0^2 + 1) + Σ(1 to ∞) 1/(k^2 + 1)
................................= 1 + 2 * Σ(k = 1 to ∞) 1/(k^2 + 1).

So, 1 + 2 * Σ(k = 1 to ∞) 1/(k^2 + 1) = π coth π
==> Σ(k = 1 to ∞) 1/(k^2 + 1) = (π coth π - 1)/2.

Double check:
http://www.wolframalpha.com/input/?i=%CE%A3%28k+%3D+1+to+%E2%88%9E%29+1%2F%28k^2+%2B+1%29+

I hope this helps!

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Evaluate the sum 1/(k^2+1) from k= 1 to infinity?

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