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.