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Using the maclaurin expansions of coshx and sinhx I came up with $e^x = \sum_{n=0}^\infty$${x^{2n}(2n+1+x)}\over {(2n+1)!}$

Plugging in $x=1$ I got:

$$e = \sum_{n=0}^\infty {2(n+1)\over (2n+1)!}$$

I can't see any way to simplify this to the more general form of $e = \sum_{n=0}^\infty \frac 1 {n!}$

Is there one?

Community
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Rohil Verma
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1 Answers1

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$$ \frac{2(n+1)}{(2n+1)!} = \frac{2n+1}{(2n+1)!} + \frac{1}{(2n+1)!} = \frac{1}{(2n)!} + \frac{1}{(2n+1)!}, $$ and then $$ \sum_{n=0}^{\infty} \left( \frac{1}{(2n)!} + \frac{1}{(2n+1)!} \right) = \sum_{k=0}^{\infty} \frac{1}{k!}. $$

Chappers
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  • Thanks a lot, but how did you go from your penultimate step to your final form? Apologies if this is trivial, I'm not very experienced. – Rohil Verma Apr 26 '15 at 15:45
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    @RohilVerma Every integer is even or odd; the sum on the left has the evens and odds separated: it can be written as $\sum_{k \text{ even}} \frac{1}{k!}+\sum_{k \text{ odd}} \frac{1}{k!}$. – Chappers Apr 26 '15 at 15:52
  • Of course, thanks so much! – Rohil Verma Apr 26 '15 at 15:53