Infinity series of 1/(2n)!

Question

What's the sum:

$$ \sum_{n=0}^\infty \frac{1}{(2n)!} $$

I tried to apply $e = x^n/n!$, but not getting the required formula. Any help?

Answer 1

You've stepped into the world of hyperbolic functions. Define $$\cosh x=\frac{\exp x+\exp -x}{2},\,\sinh x:=\frac{\exp x-\exp -x}{2},$$the even and odd parts of $\exp x$. (The name "hyperbolic" reflects the fact that $\cosh^2x-\sinh^2x=1$, so $u=\cosh x,\,v=\sinh x$ parameterises the $u>0$ part of the hyperbola $u^2-v^2=1$.) Your sum is just $$\cosh 1=\frac{e^2+1}{2e}.$$

Answer 2

$e^x =1+x/1!+x^2/2! +x^3/3!+..,$

abs. convergent for $x \in \mathbb{R}$.

$S_n := \sum_{0}^{n}(1)^n/n!$

$T_n := \sum_{0}^{n}(-1)^n/n!$

$R_n := (1/2)(S_n+T_n) =$

$(1/2)\sum_{0}^{n}\dfrac{1+(-1)^n}{n!}$.

$\lim_{n \rightarrow \infty}R_n=$

$(1/2)\lim_{n \rightarrow \infty}(S_n+T_n)=$

$(1/2)(\lim_{n \rightarrow \infty}S_n + \lim_{n \rightarrow \infty}T_n)$.

Finally:

$\lim_{n \rightarrow \infty} \sum_{0}^{\infty}\dfrac{1}{(2n)!}= (1/2)(e^1+e^{-1}).$

Answer 3

Here's another easy way I found:

We know the taylor series: $e^x = \sum_{n=0}^\infty \frac{x^n}{n!} $

We can split this series up into the even and odd terms as follows:

$ e^x = \sum \frac{x^{2n}}{(2n)!} + \sum \frac{x^{2n+1}}{(2n+1)! } $

Set $x=1$: $e^1 = \sum \left( \frac{1}{(2n)!} + \frac{1}{(2n+1)!} \right) $ (1)

Set $x=-1$: $e^1 = \sum \left( \frac{1}{(2n)!} - \frac{1}{(2n+1)!} \right) $ (2)

Add (1) and (2): $ e + e^{-1} = 2 \sum \frac{1}{(2n)!} $

Subtract (1) and (2): $ e - e^{-1} = 2 \sum \frac{1}{(2n+1)!} $

Or: $ \sum \frac{1}{(2n)!} = \frac{e + e^{-1}}{2} $ and $ \sum \frac{1}{(2n+1)!} = \frac{e - e^{-1}}{2} $

which is what we needed to find.