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I want to calculate the infinite sum: $$\sum^{\infty}_{k=1} \frac{e^{-5}5^{2k-1}}{(2k-1)!}$$ I know this series converges by the ratio test. So I must compute the limit: $$\lim_{n \to \infty} \sum^{n}_{k=1} \frac{e^{-5}5^{2k-1}}{(2k-1)!}$$.

Now I can't spot any links with this summation, how would I evaluate it?

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1 Answer 1

up vote 5 down vote accepted

Use the fact that

$$\sinh{x} = \frac12 \left (e^x-e^{-x} \right ) = \sum_{k=1}^{\infty} \frac{x^{2 k-1}}{(2 k-1)!}$$

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ah nice.+1 ${}{}{}$ – Sabyasachi Mar 16 '14 at 13:39
@Ron Gordon. You are too fast for me ! Cheers. – Claude Leibovici Mar 16 '14 at 13:39
Is this it's taylor expansion? – user2850514 Mar 16 '14 at 13:39
@user2850514: yes indeed. – Ron Gordon Mar 16 '14 at 13:40

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