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 $0<p\leq1$. 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?
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1$\begingroup$ could express it as a double summation and try to manipulate then change the order? $\endgroup$– Henry LeeAug 14, 2021 at 13:33
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$\begingroup$ I am not sure I understand what you are saying.. Which is going to be the second' s summation bound? $\endgroup$– Νικολέτα ΣεβαστούAug 14, 2021 at 13:37
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$\begingroup$ $$ e^{n^p } = 1 + n^p + \frac{(n^p)^2}{2!} +\cdots + \frac{{(n^p )^{\left\lceil {2/p} \right\rceil } }}{{\left\lceil {2/p} \right\rceil !}} + \cdots \ge \frac{{(n^p )^{\left\lceil {2/p} \right\rceil } }}{{\left\lceil {2/p} \right\rceil !}} \ge \frac{{n^2 }}{{\left\lceil {2/p} \right\rceil !}} $$ $\endgroup$– GaryAug 14, 2021 at 13:51
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2$\begingroup$ Prove is the verb and proof is the noun. $\endgroup$– vitamin dAug 14, 2021 at 13:59
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$\begingroup$ I would say, for any $p>0$, prove $$\lim_{n\to\infty}\frac{e^{-n^p}}{1/n^2} = 0$$ $\endgroup$– GEdgarAug 15, 2021 at 0:53
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2 Answers
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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 $\sum_{n}1/n^{pm}<\infty$.
answered Aug 14, 2021 at 13:36
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$\begingroup$ Thanks I wouldn't have thought it. $\endgroup$ Aug 14, 2021 at 13:50
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Do you see , $e^x>x^k/k!\,,\, \forall k\in \mathbb{N},x\in\mathbb{R}^+$?
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$\begingroup$ 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). $\endgroup$ Aug 14, 2021 at 13:58
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$\begingroup$ I'll change it to line up with the comments $\endgroup$ Aug 15, 2021 at 0:30