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

Question

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

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 06:43
@Hans: Thanks for that. An extremely useful clarification for me. — NotSuper, Nov 30 '10 at 07:23
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
See also https://math.stackexchange.com/questions/1708900/sum-of-sum-limits-n-0-infty-frac1kn — Watson, Oct 31 '16 at 17:52

score 36 · Accepted Answer · answered Nov 29 '10 at 12:12

36

It's half the sum of $e^1=\sum 1/n!$ and $e^{-1}=\sum (-1)^{n}/n!$ (or $\cosh 1$, in other words).

answered Nov 29 '10 at 12:12

Hans Lundmark

53,395

2

@Hans: Tell me a bit more? Why is it that? Thanks. – NotSuper Nov 29 '10 at 13:24
11

@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
@Américo: Thank you! I had to run to a seminar so I didn't have time to write a more detailed anwer... – Hans Lundmark Nov 29 '10 at 14:15
By the way, if one has found a number to high precision, one can try to see if it's known by the Inverse Symbolic Calculator (which in this case indeed answers with cosh(1)): http://oldweb.cecm.sfu.ca/cgi-bin/isc/lookup?number=1.543080634815243778477905620757&lookup_type=simple – Hans Lundmark Nov 29 '10 at 14:20
Typos in my 3rd and 4th sums: should be $(2n)!$ instead of $2n!$. – Américo Tavares Nov 29 '10 at 14:24
@Americo: There seems to be something wrong -- I see only what is to me strange code. – NotSuper Nov 29 '10 at 15:45
Ah, you guys are amazing. Now, rather than asking for a proof (although I'm glad to see it), I wanted to know how you knew that my series could be taken apart into those 2 pieces, and that the limit is (e + 1/e)/2. – NotSuper Nov 29 '10 at 16:42
4

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
Well, I have to confess that I don't even know what cosh(x) means. – NotSuper Nov 29 '10 at 17:01
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It's just an abbreviation for $(e^x+e^{-x})/2$. See http://en.wikipedia.org/wiki/Hyperbolic_function. – Hans Lundmark Nov 29 '10 at 17:02
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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: http://math.stackexchange.com/questions/11739/summing-up-the-series-a-3k-where-log1-xx2-sum-a-k-xk/11741#11741. – Hans Lundmark Nov 29 '10 at 17:06
But now I sort of do, because I just read the Wikipedia article http://en.wikipedia.org/wiki/Hyperbolic_function . Also Hans, thanks for the link to Inverse Symbolic Calculator. Very interesting. – NotSuper Nov 29 '10 at 17:08
@Hans: Ha! I see our comments crossed each other. I'll look at that other question. – NotSuper Nov 29 '10 at 17:11

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

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