Can series start with $-∞$? So, I decided to dig a little deeper into numerical integration because we hardly had any of that in my analysis class. I've come across this method for improper integrals:
Метод Самокиша (not in English, unfortunately).
What scares me though, is that the series in the last formula starts with $- \infty$. Is that even possible? We haven't studied series yet, but from what I understand the series usually starts with $1$ (or $0$) and goes to infinity.
I really hope you can help me figure this out. Thanks!
 A: There are in fact nice examples of doubly infinite series occurring in practice. The Jacobi theta functions can be defined as doubly infinite Fourier series, e.g.
$$\vartheta_2(z,q)=\sum_{n\in\mathbb Z} q^{\left(n+\frac12\right)^2}\exp((2n+1)iz)$$
A: 
What scares me though, is that the series in the last formula starts with $−\infty$. Is that even possible? We haven't studied series yet, but from what I understand the series usually starts with $1$ (or $0$) and goes to infinity.

You should not be scared.
Series of the type:
$$\sum_{n=-\infty}^\infty a_n \qquad \text{(also denoted by } \textstyle\sum_{n\in \mathbb{Z}} a_n\text{)}$$
are usually called bilateral series. A bilateral series converges iff the limit:
$$\lim_{N,M\to \infty} \sum_{n=-M}^N a_n$$
exists; otherwise, it is said to diverge. If you want, you can think a convergent bilateral series as a sum of two "standard" series, i.e.:
$$\sum_{n=-\infty}^\infty a_n = \sum_{n=0}^\infty a_n +\sum_{n=1}^\infty a_{-n}$$
For example, the bilateral series:
$$\sum_{n=-\infty}^\infty \frac{1}{(2n+1)^2}$$
converges: in fact, for fixed $N,M\in \mathbb{N}$ you get:
$$\begin{split} \lim_{N,M\to \infty} \sum_{n=-M}^N \frac{1}{(2n+1)^2} &= \lim_{N\to \infty} \sum_{n=0}^N \frac{1}{(2n+1)^2} +\lim_{M\to \infty} \sum_{n=1}^M \frac{1}{(1-2n)^2} \\ &= \sum_{n=0}^\infty \frac{1}{(2n+1)^2} +\sum_{n=1}^\infty \frac{1}{(2n-1)^2}\end{split}$$
for both series $\sum 1/(2n+1)^2$ and $\sum 1/(1-2n)^2$ converge; in particular:
$$\sum_{n=-\infty}^\infty \frac{1}{(2n+1)^2} =\frac{\pi^2}{4}\; .$$
A: It is possible.   A way to conceptually verify this is to consider a function whose integral from -infinity to +infinity converges such as probability density functions (I.e. normal curves) whose aforementioned integral converges to one.
A: We have the following definition:
$$ \sum_{n=0}^\infty a_n = \lim_{N \to \infty} \sum_{n=0}^N a_n,$$
if this limit exists. It is now clear how to make sense of a sum which is infinite on both sides:
$$ \sum_{n=-\infty}^\infty a_n = \lim_{N\to \infty} \lim_{M \to -\infty} \sum_M^N a_n, $$
if this limit exists. Moreover, if this latter limit exists, then it is also equal to 
$$ \lim_{N\to\infty}\sum_{n=-N}^N a_n,$$
which you can evaluate numerically in the usual way -- sum up more and more terms and keep track of how small the summands are getting...
A: For conditional convergence, you're against the wall and fighting with however you decide to define a doubly infinte series I'm afraid.
