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: $\sum_{n=-\infty}^\infty \frac{1}{(2n+1)^2} =\frac{\pi^2}{4}\; .$