Example of summing an infinite series using Poisson summation I've recently come across Poisson summation several times. Although I think I understand the basic idea pretty well, I would like to see an example or two of how to use it to sum an infinite series. A lot of googling hasn't turned up any examples, but lots of related stuff on MO. If anyone knows of simple examples please share them.
The sum that got me interested in the first place is:
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\mathrm{erf}(\alpha n)$$
This is the alternating harmonic series "dressed" by the error function ($\alpha$ is just a positive constant to control how quickly the error function turns on). The corresponding Fourier series should converge rapidly.
 A: OK, Here's a simple example. Perform the sum:
$$
S = \sum_{n=1}^\infty \frac{(-1)^n}{2n-1}=-\frac{\pi}{4}
$$
using Poisson summation. Just summing terms in order is impractical; reaching machine precision would require summing $\sim$10$^{15}$ terms, requiring $\sim$4000 cpu years on my laptop.
Convergence for brute force summing (I didn't actually compute the last term in the graph [obviously!]---I just extrapolated.)
Here are the steps I followed to use Poisson summation for this sum.


*

*Extend the sum to "full range" ($n$ ranges from $-\infty$ to $+\infty$).

*Find a continuous function, $f(x)$ that matches the sum for integer arguments.

*"Periodize" $f(x)$ to make a function, $g(x)$, that is periodic from $0<x<1$ (and any other integer interval).

*Write $g(x)$ as a Fourier series.

*Evaluate at $x=0$ to recover a sum over integers.

*Evaluate the resulting sum in reciprocal space, rather than real space.


Here we go:
Step 1: Note that 
$$
2S = 2\sum_{n=1}^\infty \frac{(-1)^n}{2n-1}=\sum_{n=-\infty}^\infty \frac{(-1)^n}{2n-1}=-\frac{\pi}{2}.
$$
(The sum is "symmetric" about $n=1/2$.)
Now we have a "full range" sum ($n$ ranges from $-\infty$ to $+\infty$) that we will replace with an $f(x)$ that matches the terms in the sum when $x\in\mathbb{Z}$.
Step 2: Find the matching function:
$f(x)=\frac{\cos(\pi x)}{(2x-1)}$
does the trick. It matches the terms of the sum at integer values of $x$.
Step 3: Periodize $f(x)$:
$$
g(x)=\sum_{n=-\infty}^{+\infty}f(x+n).
$$
This function is periodic over the interval $0\leq x<1$ (or any other unit interval).
Step 4: Write $g(x)$ as a Fourier series:
$$
g(x) = \sum_{k=-\infty}^{+\infty}a_k e^{2\pi i k x} =\sum_{k=-\infty}^{+\infty} e^{2\pi i k x}\int\limits_{0}^{1}g(x')e^{-2\pi i k x'}dx= \sum_{k=-\infty}^{+\infty} e^{2\pi i k x}\int\limits_{-\infty}^{+\infty}f(x')e^{-2\pi i k x'}dx'.
$$
(It might not be obvious that $\int\limits_{0}^{1}g(x')e^{-2\pi i k x'}dx= \int\limits_{-\infty}^{+\infty}f(x')e^{-2\pi i k x}dx'$ but remember that $g(x)$ is just a sum of shifted copies of $f(x)$. Adding up those shifted copies and integrating over the unit interval is the same thing as integrating $f(x)$ over the entire domain. Cute.)
Step 5: Evaluate the Fourier series at $x=0$ (and recover a sum over integers).
$$
g(0) = \sum_{n=-\infty}^{+\infty}f(n)= \sum_{k=-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}f(x')e^{-2\pi i k x'}dx'.
$$
$$
\sum_{n=-\infty}^{+\infty}f(n)= \sum_{k=-\infty}^{+\infty}a_k.
$$
Step 6: Instead of evaluating the LHS (we know that converges slowly), evaluate the RHS.
$$
{a_k = }\frac{1}{2} \left(2 \log (\left| 1-2 k\right|  \left| 2 \pi  k+\pi \right| )-\log \left((1-2 k)^2\right)-\log \left((2 \pi  k+\pi )^2\right)-\Gamma \left(0,\frac{1}{4} (1-2 k)^2 \pi ^2\right)-\Gamma \left(0,\frac{1}{4} (2 \pi  k+\pi )^2\right)\right)
$$
(I did the integral for $a_k$ with Mathematica. $\Gamma(a,z)$ is the incomplete Gamma function.) 
$a_k=0$ for all $k$(!) except $k=0$ for which it is $-\frac{\pi}{2}$, the exact answer for $2S$.
The LHS requires $\sim$10$^{15}$ terms for machine precision. The RHS is exact using only the first term. Neat. That's fast convergence.
A: Many worked out examples are given in the relatively recent paper by Glasser and Bargis.
