Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

To put it simply: does the series $\sum_{n=1}^{\infty} \frac{n(-1)^{n}}{(2n+1)} = -\frac{1}{3} + \frac{2}{5} - \frac{3}{7} + \cdots$ converge?

share|cite|improve this question
The terms do not converge to zero, so it does not converge. If you take the terms in pairs, it does converge, but so does $\sum (-1)^n$. – Thomas Andrews Nov 14 '12 at 17:33
up vote 0 down vote accepted

Hint: Note that the absolute value of the terms does not go to zero, so it fails the alternating series test. Since the absolute value of the terms is greater than $\frac 14$, what limit and $N$ would you select if I give you $\epsilon=\frac 18?$

share|cite|improve this answer
To be honest I have no idea. I am now taking the first math courses at my university, and feel quite lost. We just began to work with the convergence of series. I think I understand limits on some level but really need very much practise. – jeee Nov 14 '12 at 17:48
@jeee: The point is that if the series converges, it needs to get close to something, which we call the limit. If the terms stay large, it bounces around too much to converge. The formal definition says that if I give you an $\epsilon \gt 0$, you can give me an $N$ so that any time I sum at least the first $N$ terms (and keep going as far as I want) it will be within $\epsilon$ of the limit. But if $\epsilon = \frac 18$ and each term is at least $\frac 14$, one more term will take you out of the allowable range. – Ross Millikan Nov 14 '12 at 17:55
Thank you for your answers. I think I get the idea now. – jeee Nov 14 '12 at 18:03

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.