# What is limit of $\sum \limits_{n=0}^{\infty}\frac{1}{(2n)!}$?

What is the limit of the series $1 \over (2n)!$ for n in $[0, \infty)$ ?

$$\sum_{n = 0}^{\infty}{1 \over (2n)!}$$ I've ground out the sum of the 1st 1000 terms to 1000 digits using Python, (see here ), but how would a mathematician calculate the limit? And what is it?

No, this isn't homework. I'm 73. Just curious.

Thanks

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A picky remark about terminology: One usually speaks of the sum of a series $\sum_0^{\infty} a_n$, not of its limit (although this value is of course defined as the limit of the sequence of partial sums $\sum_0^N a_n$ as $N \to \infty$). – Hans Lundmark Nov 30 '10 at 6:43
@Hans: Thanks for that. An extremely useful clarification for me. – NotSuper Nov 30 '10 at 7:23
+1 for still being interested in math...keep on thinking! – draks ... Jul 31 '12 at 9:06
Another interesting sum is $\sum_{0}^{\infty}\frac{1}{(3n)!} = \frac{1}{3}\left(e+2e^{-1/2}\cos\left(\frac{\sqrt{3}}{2}\right)\right)$ – SomeStrangeUser Mar 11 '14 at 13:46

It's half the sum of $e^1=\sum 1/n!$ and $e^{-1}=\sum (-1)^{n}/n!$ (or $\cosh 1$, in other words).
@NotSuper: Observe that $$\begin{eqnarray*} \sum_{n=1}^{\infty }\frac{1}{n!}+\sum \frac{\left( -1\right) ^{n}}{n!} &=&% \frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{\left( n-1\right) !}+\frac{1}{n!}% +\cdots \\ &&-\frac{1}{1!}+\frac{1}{2!}-\cdots +\frac{\left( -1\right) ^{n-1}}{\left( n-1\right) !}+\frac{\left( -1\right) ^{n}}{n!}+\cdots \\ &=&2\times \frac{1}{2!}+2\times \frac{1}{4!}+\cdots +2\frac{1}{2n!}+\cdots \\ &=&2\left( \frac{1}{2!}+\frac{1}{4!}+\cdots +\frac{1}{2n!}+\cdots \right) \\ &=&2\sum_{n=1}^{\infty }\frac{1}{\left( 2n\right) !} \end{eqnarray*}$$ – Américo Tavares Nov 29 '10 at 13:53
It's not too hard if you're familiar with the series $\cosh x=1+x^2/2!+x^4/4!+\dots$; just set $x=1$. :-) If you know the series for $\sin x$ and $\cos x$, you will easily remember the series for $\sinh x$ and $\cosh x$ just by changing all the minus signs into plus. – Hans Lundmark Nov 29 '10 at 16:49
It's just an abbreviation for $(e^x+e^{-x})/2$. See en.wikipedia.org/wiki/Hyperbolic_function. – Hans Lundmark Nov 29 '10 at 17:02
By the way, the construction $(f(x)+f(-x))/2$ will automatically pick out the even-numbered terms in the power series for $f(x)$. See this other question for how to pick out every third term: math.stackexchange.com/questions/11739/…. – Hans Lundmark Nov 29 '10 at 17:06