$\sum_{n=0}^\infty \frac{1}{(2n)!}=\cosh(1)$ Is there any way to show that $$\sum_{n=0}^\infty \frac{1}{(2n)!}=\cosh(1) = \frac{1+e^2}{2e}$$ knowing that $$\sum_{n=0}^\infty \frac{1}{n!}=e$$
I guess that we can also show that $\sum_{n=0}^\infty \frac{1}{(2n-1)!}=\sinh(1) $ but I have no clue of how to get rid of the "2" in the factorial...

Write down the series expansions for $e^x$ and $e^{-x}$ around $0$. Add the two series and divide by $2$. Similarly, you can subtract them, to get $\sinh$.
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\\e^{-x}=\sum_{n=0}^\infty\frac{(-1)^nx^n}{n!}$$
When you add these two expressions, the odd terms will cancel. If you subtract them, the even terms will cancel out.
