Question

Find the sum of the infinite series $1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+....$

Answer 1

Fact: $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

Thus $e^1=e=1+1+\frac{1}{2!}+\frac{1}{3!}+....$ ---$\left(1\right)$, (substitute $x=1$)

and $e^{-1}=\frac{1}{e}=1-1+\frac{1}{2!}-\frac{1}{3!}+...$ ---$\left(2\right)$, (substitute $x=-1$)

Now add $\left(1\right)$ and $\left(2\right)$

$\Rightarrow e+\frac{1}{e}=2(1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+.........)$

$\therefore 1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+....=\frac{1}{2}(e+\frac{1}{e})$