I want to show that $$\cosh(x) = \sum_{n=0}^{\infty} \frac{(x)^{2n}}{(2n)!} $$
I know that $cosh(x) = \frac{exp(x)+exp(-x)}{2}$ but i cant seem to get there from the original series. I know that $$ exp(x) = \sum_{n=0}^{\infty} \frac{(x)^{n}}{(n)!}$$
and i split up the $x^{2n}$ into $x^n x^n$ but i can't seem to get the negative exp or simplify the (2n)!. I basically want to show that the series is cosh(x).
The trick is to play around with the Taylor Series terms of $e^x$. $$\cosh(x) = \frac{e^{x}+e^{-x}}{2} \\ = \frac{\sum_{n=0}^\infty\frac{x^n}{n!}+\sum_{n=0}^\infty\frac{(-x)^n}{n!}}{2}$$ Now break the sums apart further into their even and odd terms, cancel the odd ones, group the even ones and see what is left. For example, you can write the sum $\sum_{i=1}^\infty i$ as $$\sum_{i=1}^\infty i= \sum_{i=1}^\infty 2i + \sum_{i=1}^\infty (2i-1)$$