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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 $N \to \infty$).2010-11-30
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    @Hans: Thanks for that. An extremely useful clarification for me.2010-11-30
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    +1 for still being interested in math...keep on thinking!2012-07-31
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    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)$2014-03-11
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    See also https://math.stackexchange.com/questions/1708900/sum-of-sum-limits-n-0-infty-frac1kn2016-10-31

1 Answers 1

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

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    @Hans: Tell me a bit more? Why is it that? Thanks.2010-11-29
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    @NotSuper: Observe that $$\begin{eqnarray*} \sum_{n=1}^{\infty }\frac{1}{n!}+\sum \frac{\left( -1\right) ^{n}}{n!} &=&% \frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{\left( n-1\right) !}+\frac{1}{n!}% +\cdots \\ &&-\frac{1}{1!}+\frac{1}{2!}-\cdots +\frac{\left( -1\right) ^{n-1}}{\left( n-1\right) !}+\frac{\left( -1\right) ^{n}}{n!}+\cdots \\ &=&2\times \frac{1}{2!}+2\times \frac{1}{4!}+\cdots +2\frac{1}{2n!}+\cdots \\ &=&2\left( \frac{1}{2!}+\frac{1}{4!}+\cdots +\frac{1}{2n!}+\cdots \right) \\ &=&2\sum_{n=1}^{\infty }\frac{1}{\left( 2n\right) !} \end{eqnarray*}$$2010-11-29
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    @Américo: Thank you! I had to run to a seminar so I didn't have time to write a more detailed anwer...2010-11-29
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    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=simple2010-11-29
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    Typos in my 3rd and 4th sums: should be $(2n)!$ instead of $2n!$.2010-11-29
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    @Americo: There seems to be something wrong -- I see only what is to me strange code.2010-11-29
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    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)/2.2010-11-29
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    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.2010-11-29
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    Well, I have to confess that I don't even know what cosh(x) means.2010-11-29
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    It's just an abbreviation for $(e^x+e^{-x})/2$. See http://en.wikipedia.org/wiki/Hyperbolic_function.2010-11-29
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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#11741.2010-11-29
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    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.2010-11-29
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    @Hans: Ha! I see our comments crossed each other. I'll look at that other question.2010-11-29