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

I wanted to find the limit of the series $$1+ \frac{1}{2!}+ \frac{1}{4!}+\dotsb$$. My approach: Let $$S$$ be the required sum.

Then $$S= (1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dotsb)- (1+ \frac{1}{3!}+...)$$

i.e., $$S= e - (1+ \frac{1}{3!}+\dotsb)$$ But I don't know how to proceed further. I want to work the problem on my own. So please give me hint rather than the whole answer.

• This is $\cosh(1) = \frac 1 2(e + 1/e)$. – user296602 Sep 21 '18 at 20:43
HINT: If we define $$G(x):=\sum_{n=0}^\infty a_n x^n$$ Then what is the series representation of $$G(x)+G(-x)=\sum_{n=0}^\infty \space ?$$
$$\frac{1^n+(-1)^n}{2}$$
Is $1$ when $n$ is even, and $0$ when $n$ is odd.