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

I need to show, using the comparison test, that $\sum\limits_{n = 1}^{+\infty} e^{-\sqrt{n + 1}}$ converges, but I can't come up with a larger convergent series.


share|cite|improve this question
You may try $\sum{\frac{1}{n^2}}$, for example, if you know that $\sqrt{n+1}$ is larger than $2\log{n}$ for sufficiently large $n$. – user20266 Jun 4 '12 at 15:07
up vote 7 down vote accepted

Since $\mathrm e^u=\sum\limits_{k=0}^{+\infty}\frac{u^k}{k!}\gt\frac{u^4}{4!}$ for every $u\gt0$, one has $\mathrm e^{-\sqrt{n+1}}\lt\frac{4!}{(n+1)^2}$ hence the series converges.

A similar argument shows that the series $\sum\limits_{n}\mathrm e^{-n^a}$ converges for every $a\gt0$.

share|cite|improve this answer
Would you mind explaining your answer in more detail? I understand $e^u = \sum\limits_{n = 0}^{+\infty} \frac{u^n}{n!}$, but $\frac{u^4}{4!}$ and $\frac{4!}{{(n + 1)}^2}$ come from? – David Robert Jones Jun 4 '12 at 15:42
In the series you understand, all the terms are positive hence one term (here $\frac{u^4}{4!}$) is less than the sum $\mathrm e^u$ of the series. Write this as $\frac{4!}{u^4}\gt\mathrm e^{-u}$ and use it for $u=\sqrt{n+1}$. – Did Jun 4 '12 at 15:46
got it! very impressive. – David Robert Jones Jun 4 '12 at 16:51

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.