A way to do this is using regularization by substracting a continuous integral, ,with the help of the Euler-MacLaurin formula:

You can write :

$$ \sum_{Regularized} =(\sum_{n=0}^{+\infty}f(n) - \int_0^{+\infty} f(t) \,dt) = \frac{1}{2}(f(\infty) + f(0)) + \sum_{k=1}^{+\infty} \frac{B_k}{k!} (f^{(k - 1)} (\infty) - f^{(k - 1)} (0))$$
where $B_k$ are the Bernoulli numbers.

With the function $f(t) = te^{-\epsilon t}$, with $\epsilon > 0$, you have $f^{(k)}(\infty) = 0$ and $f(0) = 0$, so with the limit $\epsilon \rightarrow 0$, you will find :
$$\sum_{Regularized} = - \frac{B_1}{1!} f (0) - \frac{B_2}{2!} f' (0) = - \frac{1}{12}$$

because $f(0) = 0$ and $B_2 = \frac{1}{6}$

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user Trimok