The Series to Ad Infinitum part I

Note that, in general, $e^{x} = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^{n}}{n!}.$

So if we add the series for $e^{x}$ and $e^{-x}$ together, all the odd-powered terms will cancel and the even-powered terms will be duplicated. Thus

$\dfrac{e^{x} + e^{-x}}{2} = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^{2n}}{(2n)!}.$

This of course is the series for $\cosh(x).$ So for this question we have $x = \dfrac{1}{2}$ , so the desired sum is

$\cosh(\dfrac{1}{2}) = \dfrac{\sqrt{e} + \dfrac{1}{\sqrt{e}}}{2} = \boxed{\dfrac{e + 1}{2\sqrt{e}}}.$

Why is this level 4? Easier than most other level 4 problems...