I can show that $\sum^\infty_{k=1}{{1}\over{k^2}} = {{\pi^2}\over{6}}$ without to much hassle just using two representations of ${{sin(x)}\over{x}}$ but I cant find any proof nearly as simple when I try to determine what $\sum^\infty_{k=1}{{1}\over{k^2+1}}$ equals.
Does there exist an elementary proof? or must I resort to using integrals?
You can evaluate a series for cotangent at the square root of -1: see
http://en.wikipedia.org/wiki/Trigonometric_functions#Series_definitions .
There is a general method for computing sums of the type in your question that DOES use integration.
Let $f(z)=P(z)/Q(z)$ where $P(z)$ and $Q(z)$ are polynomials satisfying $\deg Q \geq 2+\deg P$. Assume that $f(z)$ has no poles at the points $0,\pm 1, \pm 2, \ldots$. Then
$\lim_{N \rightarrow \infty}\sum_{k=-N}^{N} f(k)$
is equal to
-{sum of residues of $\pi f(z) \cot(\pi z)$ at the poles of $f(z)$}.
The identity is proved using Cauchy's residue theorem and the square contour with corners at $(N+1/2)(1+i)$, $(N+1/2)(-1+i)$, $(N+1/2)(-1-i)$, $(N+1/2)(1-i)$. For a reference, see the exercises in section 6.3 of Saff and Snider's "Fundamentals of Complex Analysis" third edition. Several examples are given.
In your question, by setting $f(z)=1/(z^2+1)$ you can derive the formula
$\sum_{k=-\infty}^{\infty}\frac{1}{k^2+1} = \pi \coth(\pi)$.
You can also compute the Fourier series of $f\left(x\right)=\begin{cases} 0 & -\pi\le x<0 \ , \ e^{x} & 0 \le x \le \pi \end{cases}$
and then substitute $x=0,\pi$.
