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

This series $\sum_{n=1}^{\infty}\frac{1}{n}\sum_{k=1}^{\infty}e^{-(n-k)^{2}}$ converges? I think not, I used the integral: $\int_{1}^{\infty}\int_{1}^{\infty}\frac{1}{x}e^{-(x-y)^{2}}dxdy\rightarrow \infty$ then the series diverges. What do you think?

share|cite|improve this question
A lot simpler than that! Look at inner sum, put $k=n$. You get $1$. So inner sum is bigger than $1$. Then it is comparison with harmonic series. – André Nicolas Mar 2 '12 at 23:42
up vote 3 down vote accepted

HINT: $$ \sum_{n=1}^m \frac{1}{n}\sum_{k=1}^\infty \exp(-(n-k)^2) > \sum_{n=1}^m \frac{1}{n} \sum_{k=1}^\infty \delta_{k,n} \exp(-(n-k)^2) = \sum_{n=1}^m \frac{1}{n} $$

share|cite|improve this answer
Upvoted! Would you mind giving this a try? – Pedro Tamaroff Mar 2 '12 at 22:42

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.