Sign up ×
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.

Let $\{a_n\}$ be a sequence of positive real numbers which diverges to $+\infty$. Then when can we say that the following series converges? $$\sum_{n\ge 1}e^{-a_n}$$ Thanks in advance.

share|cite|improve this question
The \large command is at best distracting. – Did Apr 26 '14 at 16:56
Okay, thanks @Did. – Samrat Mukhopadhyay Apr 26 '14 at 16:59

1 Answer 1

Nothing: it may converge or diverge:

  • divergence: $a_n:=\log n$;
  • convergence: $a_n=2\log n$.

The updated question reduces to the following: if $b_n$ is a sequence of positive numbers converging to $0$, when does the series $\sum_{n\geqslant 1}b_n$ converge, which is too broad.

share|cite|improve this answer
I think you may tell him the range of $a_n$ for which it diverges and for which it converges. – William Hilbert Apr 26 '14 at 16:49
@WilliamHilbert And how would one define this "range"? – Did Apr 26 '14 at 16:50
@Did: It seems to me that the series converges for all $a_n$ $\geq$ $2 \ln n$. Probably it's enough to answer your question. – William Hilbert Apr 26 '14 at 16:52
Is it possible to tell when the series will converge even if the sequence $a_n$ diverges? Maybe I'll update in my question to make it clearer. – Samrat Mukhopadhyay Apr 26 '14 at 16:52
@WilliamHilbert This can hardly define a "range" since the series converges for some sequences such that $a_n\lt2\log n$ for every $n$. Even more annoyingly, it also converges for some sequences such that $a_n\lt\log n$ for infinitely many $n$... – Did Apr 26 '14 at 16:55

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.