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I need to examine convergence of the following sum:

$$\displaystyle \sum^{\infty}_{n=2}\frac{n!(1+\frac{1}{2n})^{n^2}}{n^n}$$

I know that: $$\displaystyle \lim_{n \to \infty}\frac{n!(1+\frac{1}{2n})^{n^2}}{n^n} = 0$$

And that all elements of sequence $a_n$ are positive. Therefore I can use Cauchy's root test: $$\sqrt[n]{\frac{n!(1+\frac{1}{2n})^{n^2}}{n^n}}= \frac{\sqrt[n]{n!}(1+\frac{1}{2n})^{n}}{n}$$ $$\sqrt[n]{\frac{\sqrt[n]{n!}(1+\frac{1}{2n})^{n}}{n}}= \frac{\sqrt[n^2]{n!}(1+\frac{1}{2n})}{\sqrt[n]{n}}$$

But it gives me noting since I don't know what happens to $\sqrt[n^2]{n!}$ when $n \to \infty$. I tried also d'alembert's ratio test, but it gives me nothing. I think it may be LTC on limits.

StubbornAtom
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mathp
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  • Stirling's Formula estimates $n!$ in terms of $n^n$. What I am curious is that I don't recall Cauchy's criterion be using root tests. Are there references for that? – macton Jan 29 '21 at 05:49
  • Oh, it's called 'root test' in English, I'm sorry. Unfortunately, I can't use Stirling's Formula, Taylor's series nor l'Hospital. – mathp Jan 29 '21 at 05:58
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    In fact the second root test is well, invalid as that is not how that works. You only do once. – macton Jan 29 '21 at 06:10
  • I think I can do it twice (as every other of those basic tests). THe problem is, every time I use it, it "cuts" some of the information comming from original sequence. – mathp Jan 29 '21 at 06:14
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    In this case, it's a big cut: the second limit is 1 (inconclusive), and the first is not (see the answer). – macton Jan 29 '21 at 06:19
  • So, you say that I can't do the same test more than once? What about merging other ones together? – mathp Jan 29 '21 at 06:21
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    Consider $$a_n = \frac{1}{2^n}$$ the root test will give you $1/2$, but if you apply this value to the root test again, you will get $1$. The problem is that, you are in fact doing the root test to the sequence $\sqrt[n]{a_n}$ on the so-called second root test, not the original sequence. – macton Jan 29 '21 at 06:26
  • And well, you can always refer to other tests if the test you used is inconclusive. – macton Jan 29 '21 at 06:32
  • Yeah, but it seams a bit tricky to me. You gave me an example that shows that it's not always correct to use root test twice. But (if you know) I would like to find out if it is always a bad idea (does it work in any case at all?). And if it does / doesn't, what about other tets? Is it ok to merge other tests with each other, or if I calculate my limit once with one test I shouldn't ever do that again? As I suppose it's not that easy and depends on particular cases, but maybe I'm wrong. – mathp Jan 29 '21 at 06:38
  • Hmm, well it depends on a lot of a different cases, sorry if I can't answer that :( – macton Jan 29 '21 at 07:42

3 Answers3

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Hint: $$\lim_{n \to \infty} \frac{\sqrt[n]{n!}(1+\frac{1}{2n})^{n}}{n} = \lim_{n \to \infty}\frac{\sqrt[n]{n!}}{n} \lim_{n \to \infty}\left(1+\frac{1}{2n}\right)^{n}$$ To justify the split of limit, you should note that $$\lim_{n \to \infty}\left(1+\frac{1}{2n}\right)^{n}$$ is a common limit (and I'm sure you can find it). For the first limit, refer to here.

macton
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An alternative method. First note that $$ \left( {1 + \frac{1}{{2n}}} \right)^{n^2 } = \left( {\left( {1 + \frac{1}{{2n}}} \right)^{2n} } \right)^{n/2} < e^{n/2} $$ for all $n\geq 1$. Second, \begin{align*} \log n! & = \sum\limits_{k = 1}^n {\log k} = \log n + \sum\limits_{k = 1}^{n - 1} {\log k} < \log n + \sum\limits_{k = 1}^{n - 1} {\int_k^{k + 1} {\log tdt} } \\ & = \log n + \int_1^n {\log tdt} = (n + 1)\log n - n + 1. \end{align*} Thus $$ n! < en\frac{{n^n }}{{e^n }} \Rightarrow \frac{{n!}}{{n^n }}\left( {1 + \frac{1}{{2n}}} \right)^{n^2 } < ene^{ - n/2} . $$ But $$ \sum\limits_{n = 1}^\infty {ne^{ - n/2} } = \frac{{\sqrt e }}{{(\sqrt e - 1)^2 }} \approx 3.9 $$ converges, so does the original series.

Gary
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Since $n!^{1/n} \approx n/e$, the n-th root is about

$(n/e)(1+1/(2n))^n/n \approx e^{1/2}/e = e^{-1/2}<1$

so the sum converges.

marty cohen
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