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It was a new contributor's question. I answered, got my -1 again and then deleted. Then I asked myself. Then gave it up again. Actually I was gonna ask a different question NOW. When I pressed ask a question, to my surprise, the question I intended to ask yesterday was in the memory!

I wanted to evaluate the following limit by logarithmic limit rule: $$\lim_{n\rightarrow\infty} \left(\frac{n^{n-1}}{(n-1)!}\right)^{\frac{1}{n}}=\exp\left(\lim_{n\rightarrow\infty}\frac{(n-1)\ln n-\ln (n-1)!}{n}\right)=\exp\left(\lim_{n\rightarrow\infty}-\frac{1}{n}\sum_{k=1}^n\ln(\frac{k}{n})\right)$$ Then I observed a Riemann sum of an indefinite integral inside so that the limit is $$\exp\left(-\int_0^1\ln xdx\right)=\exp\left((x-x\ln x\vert_0^1)\right)=e.$$ Is my solution correct? Can you suggest another way? Stirling's approximation formula is excluded.

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    $\begingroup$ Instead of the n-root take the ratio of consecutive terms. You get the result much easier. This is pretty standard think to do and is related to root test and ratio test for series. $\endgroup$
    – Mittens
    Commented Nov 17, 2022 at 21:11
  • $\begingroup$ @OliverDíaz Another way for this limit? $\endgroup$
    – Bob Dobbs
    Commented Nov 17, 2022 at 21:18
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    $\begingroup$ This is a fine way to do it, modulo the fact that $\ln x$ goes to $-\infty$ as $x\to 0^+,$ so simple Riemann sums don't always converge to the integral. Essentially, you have the "lower" Riemann sum converges to your limit, but the "upper" Riemann sum is undefined. So you will need some fancy footwork to get the argument to work. If you know Stirling's approximation, you could use that instead. en.wikipedia.org/wiki/Stirling's_approximation?wprov=sfti1 $\endgroup$ Commented Nov 17, 2022 at 21:22
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    $\begingroup$ See math.stackexchange.com/q/1885090/42969 for a rigorous argument that $\frac{1}{n}\sum_{k=1}^n\ln(\frac{k}{n})$ converges to $\int_0^1 \ln(x) dx$. $\endgroup$
    – Martin R
    Commented Nov 17, 2022 at 21:29

2 Answers 2

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It is correct up to two things.

First, you should not write $\lim_{n \to +\infty}$ before the existence of the limit is proved. You have to start with $$\Big(\frac{n^{n-1}}{(n-1)!}\Big)^{1/n} = \Big(\frac{n^{n}}{n!}\Big)^{1/n} = \exp\Big(n \ln(n) - \frac{1}{n} \sum_{k=1}^n \ln(k)\Big) = \exp\Big(- \frac{1}{n}\sum_{k=1}^n \ln(k/n)\Big).$$

Second, the theorem form the convergence of Riemann sums applies to continuous functions on $[0,1]$. Here, the function $\ln$ is continuous on $]0,1]$ only. Yet, it is increasing, so you can add inequalities $$\int_{(k-1)/n}^{k/n} \ln x dx \le \frac{1}{n} \ln(k/n) \le \int_{k/n}^{(k+1)/n} \ln x dx.$$ over all $k \in \{1,\ldots,n\}$, to get a lower bound and an upper bound and prove the existence of the limit.

Other method : $$\frac{(k+1)^{k+1}/(k+1)!}{k^k/k!} = \frac{(k+1)^k}{k^k} = (1+1/k)^k \to e \text{ as } k \to +\infty,$$ so an application of Ces`ar'o lemma (take logarithms) yields $$\Big(\frac{n^{n}}{n!}\Big)^{1/n} \to e \text{ as } k \to +\infty.$$

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  • $\begingroup$ I am EEE engineer. Sorry. $\endgroup$
    – Bob Dobbs
    Commented Nov 17, 2022 at 21:46
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By the Stirling formula: $$\dfrac{n^{n-1}}{(n - 1)!} = \dfrac{n^n}{n!} \sim \dfrac{e^n}{\sqrt{2 \pi n}}$$ then : $$\left(\dfrac{n^{n-1}}{(n - 1)!}\right)^{\frac{1}{n}} \sim \dfrac{e}{(2 \pi n)^{1/2n}} \to e$$

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  • $\begingroup$ Nice! But Stirling's formula is exluded. Thanks. $\endgroup$
    – Bob Dobbs
    Commented Nov 17, 2022 at 21:38
  • $\begingroup$ I don't give -1. $\endgroup$
    – Bob Dobbs
    Commented Nov 17, 2022 at 21:48

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