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Does the following series converge or diverge? $$\sum_{n=1}^{\infty}e^{-n+\sqrt{n}} n^4$$

I tried using the root test to prove convergence. So we take $\quad\lim_{n\rightarrow\infty}\sqrt[n]{a_n}\quad$, but I don't know how to find this limit. I tried with derivatives but in the end I can't find a solution. Could someone please help me?

222111
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    Wolfram Alpha says 241.1710913 – Ami Mar 28 '23 at 01:28
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    It will converge. The handwavy explanation is that $e^{-n}$ dominates everything else pictured. A less handwavy explanation might have you compare to more wellbehaved series, for example rather than the term $e^{-n+\sqrt{n}}$ could you compare to $e^{-n+\frac{1}{2}n}$? To handle the fact that there is the $n^4$ there, can you make an argument about derivatives or the like, or as you suggest... using a test like root tests or ratio tests? – JMoravitz Mar 28 '23 at 01:28
  • I think the ratio test might prove a little easier to apply here. – Robert Shore Mar 28 '23 at 02:13
  • @AmiAshman: And as always the Mathematica group is off by a factor of 5. Lordy! – A rural reader Mar 28 '23 at 03:51

2 Answers2

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Let us rewrite the term a little bit. If we apply the $n$-th root, we get

$$(e^{-n + \sqrt n} n^4)^\frac{1}{n} = (e^{-1 + \frac{1}{\sqrt n}}) \cdot n^\frac{1}{n} \cdot n^\frac{1}{n} \cdot n^\frac{1}{n}\cdot n^\frac{1}{n}.$$

First, it is well-known that $ n^\frac{1}{n}$ tends to 1 for $n \to \infty$. Second, since the exponential function is continues, we get $$\lim_{n \to \infty} \exp\left({-1 + \frac{1}{\sqrt n}}\right) = \exp\left( \lim_{n \to \infty} {-1 + \frac{1}{\sqrt n}}\right) = \exp(-1),$$ where we used that $\frac{1}{\sqrt n}$ converges to 0. Now, we combine these two facts with the product rule of limits (i.e. $\lim a_n \cdot b_n = \lim a_n \cdot \lim b_n$ if both limits exist) to get $$\lim_{n \to \infty} (e^{-1 + \frac{1}{\sqrt n}}) \cdot n^\frac{1}{n} \cdot n^\frac{1}{n} \cdot n^\frac{1}{n}\cdot n^\frac{1}{n} = \exp(-1) \cdot 1 \cdot 1 \cdot 1 \cdot 1 = \exp(-1) < 1$$ Thus, we can use the root test to show convergence.

win8789
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You could use the ratio test $$a_n=e^{-n+\sqrt{n}}\, n^4 \qquad \implies \log(a_n)=-n+\sqrt{n}+4\log(n)$$ $$\log(a_{n+1})-\log(a_n)=\sqrt{n+1}-\sqrt{n}+4 \log (n+1)-4 \log (n)-1$$ Using Taylor series $$\log(a_{n+1})-\log(a_n)=-1+\frac{1}{2 \sqrt{n}}+O\left(\frac{1}{n}\right)$$ $$\frac{a_{n+1}}{a_n}=e^{\log(a_{n+1})-\log(a_n)}=\frac 1 e+\frac{1}{2 e\sqrt{n}}+O\left(\frac{1}{n}\right)$$

Claude Leibovici
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