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I am a highschool teacher and will teach integrals next term. Today I saw a video from blackpenredpen in which he computed the limit $\displaystyle \lim_{n \to +\infty}\left(\frac{n!}{n^n}\right)^{1/n}$ using the integral of $\ln x$:

$$\begin{align} \lim_{n \to +\infty}\left(\frac{n!}{n^n}\right)^{1/n} &= \lim_{n \to +\infty}e^{\frac1n \ln\left(\frac{n!}{n^n}\right)}\\ \end{align}$$

So we can consider the limit $\displaystyle \lim_{n \to +\infty} \frac1n \ln\left(\frac{n!}{n^n}\right)$. Note that this limit equals $$\lim_{n \to +\infty} \frac1n \left(\ln \left(\frac1n\right) + \ln \left(\frac2n\right) + \ldots + \ln \left(\frac nn\right)\right).$$ This is exactly the definition of $\displaystyle \int_0^1 \ln x dx = -1$, hence the initial limit equals $e^{-1} = \frac 1e$.

This technique is not present in the course notes I'm using and I would like to include this for some of my students who work faster through my notes.

Question: do you know any other limits which are easily solved using integrals (and are much harder to solve using more standard techniques)? Any suggestions are appreciated. If possible, post only the limit itself, so I can have a go at it myself :)

I am aware of this question but am looking for limits without summations in the problemstatement.

Student
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  • While your last paragraph prohibits Riemann sums such as $\lim_{n\to\infty}\tfrac1n\sum_{k=1}^nf\left(\tfrac{k}{n}\right)=\int_0^1fdx$, these can be exponentiated to products instead. In fact, your discussed example is just $f(x):=\ln x$. So did you mean to say you want examples other than e.g. $\lim_{n\to\infty}\left(\prod_{k=1}^nf\left(\tfrac{k}{n}\right)\right)^{1/n}=\exp\int_0^1fdx$? – J.G. Nov 12 '21 at 21:05
  • English is not my native language, so I might have expressed myself badly. What I meant was that I'm looking for limits that look like 'regular' limits. The ones you could encounter in a first course on limits, but more near the end. Limits that don't look like you would solve them using integrals. The example of blackpen took me by surprise, since it was an approach I didn't expect. I hope I somewhat made clear what I meant with that last paragraph. – Student Nov 12 '21 at 21:13

2 Answers2

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Try this: $$ \lim_{n\to+\infty} {2n \choose n}^{\frac{1}{n}} $$

aliberro
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  • Good choice. As with the original factorial-based problem, there's a Stirling approximation option here that's more high-level than the integrals it will hopefully encourage. – J.G. Nov 12 '21 at 21:28
  • That one looks great. I'm teaching them about combinatorial problems at this moment. Thank you! @J.G. I don't know about Stirling approximation. I'm teaching 17-18 yo, so integration is pretty basic (up to substitution, maybe splitting in partial fractions,...). – Student Nov 12 '21 at 21:31
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    @Student $n!\sim\sqrt{2\pi n}(n/e)^n$, in case you're curious. More generally, $\lim_{n\to\infty}\binom{(a+b)^n}{an}^{1/n}$ will do. – J.G. Nov 12 '21 at 21:35
  • @aliberro: would you happen to have a hint on how to approach this problem? I tried using $\ln$ to turn the product in a sum, but I'm struggling how to group the terms and which integral to invoke... – Student Nov 12 '21 at 21:41
  • @Student you're on the right track, continue with the ln, you will eventually end up with a sum of ln(n+k over something, I won't say more, I'll leave you trying. – aliberro Nov 12 '21 at 21:43
  • So I wrote $\binom{2n}{n}$ as $\frac{2n}{n} \cdot \frac{2n-1}{n} \cdot \frac{2n-2}{n-1} \cdot \frac{2n-3}{n-1} \cdot \ldots \cdot \frac{2}{1} \cdot \frac 11$. This results (after taking $\ln$) in $n \cdot \ln 2 + \ln \left(2 - \frac 1 n\right) + \ln \left(2 - \frac{1}{n-1}\right) + \ldots + \ldots \ln 1$. I'm trying to find the integral that corresponds to that last sum. Or am I way off? – Student Nov 12 '21 at 21:59
  • It would be better if you could make it look like this (n+1)/1, (n+2)/2, (n+3)/3, ... – aliberro Nov 12 '21 at 22:39
  • Thanks for the hint! I think I found the answer :) – Student Nov 13 '21 at 14:11
  • Can I ask you where you found this example? – Student Nov 13 '21 at 14:36
  • @Student it's a typical one, if you like, start with the integral then the series, that's how most of us would do it – aliberro Nov 13 '21 at 14:50
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For example, $$\lim_{n\to\infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\dots+\frac{n}{n^2+n^2}\right) $$ With some manipulations of this expression, you can show the limit is equal to $$\int_0^1\frac{dx}{1+x^2} $$

bjorn93
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