Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 5 down vote accepted

Use the fact that

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.