question about the cosh function I am having a hard time finding the following sum:
$$\sum_{n=0}^{\infty} \frac{5^n}{(2n)!}$$
I know it has something to do with cosine since it also has $(2n)!$ in the denominator and it might also be connected with $e^x$ since the terms look very similar.
I have tried playing with algebraic manipulations but again it did not help.
Any help with starting would be appreciated
 A: $$\cosh(x) = \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}$$
Therefore,
$$\sum_{k=0}^\infty \frac{5^k}{(2k)!} = \sum_{k=0}^\infty \frac{\sqrt5^{2k}}{(2k)!} = \cosh(\sqrt5) = \frac{e^\sqrt5+e^{-\sqrt5}}{2}$$
EDIT- Let 
$$f(x) = \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}$$
$$g(x) = \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}$$
Then $f$ is even, $g$ is odd, and $f(x)+g(x)=e^x$. Note that $f$ and $g$ are defined on all of $\mathbb{R}$. Then
$$e^x+e^{-x} = (f(x)+g(x))+(f(x)-g(x)) = 2f(x)$$
Therefore,
$$f(x) = \frac{e^x+e^{-x}}{2}$$
A: HINT:
As $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$
$$e^{-y}+e^y=?$$
Can you recognize $y$ here?
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
\sum_{n = 0}^{\infty}{5^{n} \over \pars{2n}!} & =
\sum_{n = 0}^{\infty}{\pars{\root{5}}^{2n} \over \pars{2n}!} =
\sum_{n = 0}^{\infty}{\pars{\root{5}}^{n} \over n!}\,{1 + \pars{-1}^{n} \over 2} \\[5mm] & =
{\sum_{n = 0}^{\infty}\pars{\root{5}}^{n}/n! +
\sum_{n = 0}^{\infty}\pars{-\root{5}}^{n}/n!\over 2} =
{\expo{\root{5}} + \expo{-\root{5}} \over 2} =
\bbx{\cosh\pars{\root{5}}}
\end{align}
