# Summation of $e^{-n^p}$ for p>0

I need to prove that $$\forall p>0$$ $$\sum _{n \geq 1} e^{-n^p}<\infty$$ I have tried many theorems that I know from analysis but I failed with all when $$0. I have prooved though that for $$p>1$$, $$e^{-n^p}\leq \frac{1}{n^p}$$ and so it converges. Any ideas for the other case?

• could express it as a double summation and try to manipulate then change the order? Aug 14, 2021 at 13:33
• I am not sure I understand what you are saying.. Which is going to be the second' s summation bound? Aug 14, 2021 at 13:37
• $$e^{n^p } = 1 + n^p + \frac{(n^p)^2}{2!} +\cdots + \frac{{(n^p )^{\left\lceil {2/p} \right\rceil } }}{{\left\lceil {2/p} \right\rceil !}} + \cdots \ge \frac{{(n^p )^{\left\lceil {2/p} \right\rceil } }}{{\left\lceil {2/p} \right\rceil !}} \ge \frac{{n^2 }}{{\left\lceil {2/p} \right\rceil !}}$$
– Gary
Aug 14, 2021 at 13:51
• Prove is the verb and proof is the noun. Aug 14, 2021 at 13:59
• I would say, for any $p>0$, prove $$\lim_{n\to\infty}\frac{e^{-n^p}}{1/n^2} = 0$$ Aug 15, 2021 at 0:53

Pick an $$m\in\mathbb{N}$$ such that $$p>1/m$$.
We know that $$e^{u}>u^{m}$$ for large $$u>0$$. Plugging in $$u=n^{p}$$ we get $$1/e^{u}<1/n^{pm}$$. Now note that $$pm>1$$ and so $$\sum_{n}1/n^{pm}<\infty$$.
Do you see , $$e^x>x^k/k!\,,\, \forall k\in \mathbb{N},x\in\mathbb{R}^+$$?
• That doesn't seem to be true for $x=-100$ and $k=2$. (But yes, the $x>0$ case is enough to get us through here). Aug 14, 2021 at 13:58
• Yes, we need $x$ to non negative Aug 14, 2021 at 14:20