Question

Show that $1+\frac{1}{2!}+\frac{1}{4!}+.......=\frac{1}{2}(e+\frac{1}{e})$.

Answer 1

We have,

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+.......\infty$.

Then $e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+......\infty$......(1)

And

$e^{-1}$$=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-......\infty$......(2).

Now adding (1) and (2) we get,

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

or, $\frac{1}{2}(e+\frac{1}{e})=(1+\frac{1}{2!}+\frac{1}{4!}+.....\infty )$