# Convergence of the series $\sum\limits_{n}e^{-a_n}$

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.

-
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

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.

-
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