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How to determine the sum $\sum_{n=0}^\infty\frac{1}{(4n)!}$ ? Do I need to somehow convert (4n)! to (2n)! or in tasks like this, should I get the (4n)! after some multiplying?

Thank you all for your time!

purgerica
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    Structure approach: 1. Note that $1^n+i^n+(-1)^n+(-i)^n=4$ if $4\mid n$ and $=0$ otherwise. 2. Hence $4$ times your series is $$\sum_{n\geqslant 0}\frac{1^n}{n!}+\sum_{n\geqslant 0}\frac{i^n}{n!}+\sum_{n\geqslant 0}\frac{(-1)^n}{n!}+\sum_{n\geqslant 0}\frac{(-i)^n}{n!}=e^1+e^i+e^{-1}+e^{-i}=\ldots$$ – Did Jun 23 '15 at 21:04
  • Answer: (1) Search this site for "(4n)!". (2) Find this question: http://math.stackexchange.com/questions/221519/identify-infinite-sum-sum-limits-n-0-infty-fracx4n4n. (3) Put $x=1$. ;-) – Hans Lundmark Jun 23 '15 at 21:04
  • Or this one, for that matter: http://math.stackexchange.com/questions/1193695/sumfunction-of-sum-n-0-infty-fracz4n4n-and-sum-n-1-inft – Hans Lundmark Jun 23 '15 at 21:06
  • Perhaps this post might be of some assistance. – Lucian Jun 24 '15 at 01:49

2 Answers2

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Hint: (a) Write down the Maclaurin series for $\cos x$; (b) Write down the Maclaurin series for $\cosh x$, that is, $\frac{e^x+e^{-x}}{2}$; (c) Look.

André Nicolas
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$cos(x) = \sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$

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

How to get (4n)! from these (2n)! and n! ?

purgerica
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