I have been given the series below and I can't figure out how to get an exact sum for it. I can't derive it because of the $2n!$ in the denominator. I don't see how it can be turned into a geometric series or arithmetic. Any ideas?
$$\sum_{n=0}^{\infty} \frac{(-1)^{n+1}{\pi}^{2n}}{2(2n)!}$$
Recall that the Maclaurin series for cosine is given by $$ \cos(\theta) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\theta^{2n}. $$ Rewriting your series, we have $$ \sum_{n=0}^{\infty} \frac{(-1)^{n+1} \pi^{2n}}{2(2n)!} = \frac{-1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)} \pi^{2n} $$ (observe that we have factored out a $-1$ from the numerator of each term, and a $2$ from the denominator of each term). Comparing this to the Maclaurin series, we have $$ \frac{-1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)} \pi^{2n} = -\frac{1}{2} \cos(\pi) = \frac{1}{2}. $$