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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.

Thanks in advance.

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

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Hint:

$$\frac{1^n+(-1)^n}{2}$$

Is $1$ when $n$ is even, and $0$ when $n$ is odd.

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