Calculus problem No.3

Consider the following Maclaurin series :

$\begin{aligned} e^x & = 1 + \frac x{1!} + \frac {x^2}{2!} + \frac {x^3}{3!} + \frac {x^4}{4!} + \frac {x^5}{5!} + \frac {x^6}{6!} + \cdots \\ e^1 & = 1 + \frac 1{1!} + \frac 1{2!} + \frac 1{3!} + \frac 1{4!} + \frac 1{5!} + \frac 1{6!} + \cdots \\ e^{-1} & = 1 - \frac 1{1!} + \frac 1{2!} - \frac 1{3!} + \frac 1{4!} - \frac 1{5!} + \frac 1{6!} + \cdots \\ \implies e + \frac 1e & = 2 + \frac 2{2!} + \frac 2{4!} + \frac 2{6!} + \cdots \\ \implies \sum_{n=1}^\infty \frac 1{(2n)!} & = \frac {e+\frac 1e}2 - 1 = \cosh 1 - 1 \approx \boxed{0.543} \end{aligned}$