# Argue for that the series $\sum_{n=1}^\infty e^{-n^2x}$ is convergent if and only if $x>0$

Consider the series $$S = \sum_{n=1}^\infty e^{-n^2x}$$ then I have to argue for that $$S$$ is convergent if and only if $$x>0$$.

As this is if and only if I think I have to assume first that S is convergent and show that this implies that $$x>0$$ but I am not sure how to. It is easy for me see that if $$x=0$$ the series is divergent but if I were to assume that S is convergent and that for a contradiction that $$x\leq 0$$ how do I proceed? And how the other way around?

Do you mind helping me?

• If $S$ is convergent then $e^{-n^2x}\to 0$ as $n\to\infty$, which happens when $x>0$, so that $-n^2x\to -\infty$ Commented May 13, 2020 at 13:22
• What is the limit of individual terms of a convergent sequence? Would these terms have that limit if $x < 0$? Commented May 13, 2020 at 13:24
• By root test the series is convergent. Commented May 13, 2020 at 13:35
• Is it possible to use that if x>0 this must mean that every tail of S is convergent. Thus S itself must be convergent. Or do I have to use contraposition? Commented May 13, 2020 at 13:51

If S is convergent then $$e^{-n^2x}\to 0$$, which it only does if $$x>0$$. If $$x\le 0$$ then $$-n^2x\ge 0$$ and hence $$\sum_{n\in\mathbb{N}} e^{-n^2x}\ge \sum_{n\in\mathbb{N}} e^{0}= \sum_{n\in\mathbb{N}}1=\infty$$. (contraposition)

• Yes, thank you! Commented May 13, 2020 at 13:30
• Is it possible to use that if $x>0$ this must mean that every tail of S is convergent. Thus S itself must be convergent. Or do I have to use contraposition? Commented May 13, 2020 at 13:43

Another way to see it is to compare the sum in question to the integral

$$\int_{1}^{\infty} e^{-(t\sqrt{x})^2} dx =\frac{1}{\sqrt{x}}\int_{\sqrt{x}}^{\infty} e^{-s^2} ds$$ Which exists only for $$x>0$$.

If $$x>0$$, then (since $$0) $$\sum\limits_{n = 1}^\infty {e^{ - n^2 x} } \le \sum\limits_{n = 1}^\infty {e^{ - nx} } = \frac{{e^{ - x} }}{{1 - e^{ - x} }} = \frac{1}{{e^x - 1}} < + \infty .$$ If $$x\leq 0$$, the terms do not tend to zero, whence the series cannot converge.

• But am I able to use that if $x>0$ this must mean that every tail of $S$ is convergent? Thus S itself must be convergent? Commented May 13, 2020 at 16:01
• You do not have to say anything about tails. The infinite sum is bounded, whence converges.
– Gary
Commented May 13, 2020 at 16:04

If the series converges $$\implies \lim_{n\to\infty} e^{-n^{2}x} = 0$$. Do you see why $$x$$ needs to be positive now?

• Yes, thank you! Commented May 13, 2020 at 13:30