# Complex analysis to evaluate series

• Nov 27th 2006, 04:42 PM
juef
Complex analysis to evaluate series
Hey all,

According to a theorem I have seen in class, we have

$\displaystyle \sum_{\substack{n=-\infty\\n \neq z_k}}^\infty f(n) = -\sum_{\textrm{$z_k$, the poles of$f(z)$}} Res\left(\frac{\pi \cos{(z\pi)}}{\sin{(z\pi)}}f(z), z_k\right)$

It can easily be seen that the theorem works for$\displaystyle f(n) = \frac{1}{n^2}$, but I've been trying to use it to find the exact value of $\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n^2+1}$ without success. The poles of fare obviously i and -i, for which the residues are respectively $\displaystyle \frac{-\pi(-1)^i\cosh{(\pi)}}{2\sinh{(\pi)}}$ and $\displaystyle \frac{-\pi(-1)^{(-i)}\cosh{(\pi)}}{2\sinh{(\pi)}}$, which implies that $\displaystyle \sum_{n=-\infty}^\infty \frac{(-1)^n}{n^2+1} = \frac{\pi\cosh{(\pi)}}{2\sinh{(\pi)}}\left( e^{\pi} + e^{-\pi} \right)$

Of course, f is an even function, so $\displaystyle \sum_{n=-\infty}^\infty \frac{(-1)^n}{n^2+1} = \sum_{n=-\infty}^{-1} \frac{(-1)^n}{n^2+1} + \sum_{n=1}^\infty \frac{(-1)^n}{n^2+1} + 1 = 2\sum_{n=1}^\infty \frac{(-1)^n}{n^2+1} + 1$.

Therefore, $\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n^2+1} = \frac{1}{2} \left( \frac{\pi\cosh{(\pi)}}{2\sinh{(\pi)}}\left( e^{\pi} + e^{-\pi} \right) - 1 \right) = 17.77673189...$

But, when I approximate the sum in Maple or MATLAB, I get another answer (which is the right one): -0.3639849730...

Does anyone have an idea of what I have done wrong?

Thank you all!!
• Nov 28th 2006, 12:44 PM
TD!
I checked your calculations and they seem correct, but I'm not convinced by the theorem.
Are you sure this holds for arbitrary f(z)? Perhaps you could state the theorem, with the conditions.
• Nov 29th 2006, 07:08 AM
juef
The only conditions I'm aware of if that $\displaystyle f(n)$ must be analytic $\displaystyle \forall n \in \mathbb{Z}$, and that $\displaystyle |f(z)| \leq \frac{constant}{|z|^2}$ when $\displaystyle |z| \rightarrow \infty$.

I still don't understand why this doesn't work, but I've just learned a new theorem which says, under the same conditions, that if we have $\displaystyle (-1)^nf(n)$ to evaluate in the series, the same formula works except that we must use $\displaystyle \frac{\pi}{\sin{(z\pi)}}f(z)$ instead of $\displaystyle \frac{\pi \cos{(z\pi)}}{\sin{(z\pi)}}f(z)$ to calculate the residue. This last formulae works for the series I wanted to evaluate initially and gives a value of $\displaystyle \frac{1}{2} \left( \frac{\pi}{\sinh{(\pi)}}-1 \right)$.

Thank you for checking my calculations... I'm still curious about why the first theorem didn't work. Maybe I'll ask my teacher about that...
• Nov 29th 2006, 08:09 AM
TD!
The first "theorem" (I wouldn't call it a theorem) only works in specific cases, as you've illustrated yourself by saying that for another form (which also falls under the general f(n), you see?), you need another calculation.