-1

$$\lim_{n\to \infty} \left(\frac{(2n)!}{n!n^n}\right)^{\frac{1}{n}}$$

Where I started:

For me, this looked like a $1^{\infty}$ form. So I did use the 'monkey-on-the-tree' kind of methodology, where we say,

'Get the monkey down and chop off the tree.'

That is, get the $\frac{1}{n}$ down, and impose the limit to the rest, subtracting one from it. It's a trick to solve these faster...

$${\frac{1}{n}} \lim_{n\to \infty} \left(\frac{(2n)!}{n!n^n}-1\right)$$

But now I am stuck. Can anyone please help me out?

Harikrishnan M
  • 243
  • 1
    What makes you think you can take part of the function whose limit you are taking, which depends on the variable of the limit, and just pull it out of the limit any which way you want? The final result of your first limit should not involve $n$ at all; but the second limit you have has a $\frac{1}{n}$ outside the limit, which means it will remain even after you compute the limit. That should tell you there is everything fundamentally wrong with going to the second limit. It just cannot be related to the first limit. Whatever that "method" is supposed to be, it isn't "just move it wherever" – Arturo Magidin Feb 10 '24 at 02:17
  • There should be alternative methods no @ArturoMagidin? I am looking for that. As of now nothing else strikes me. – Harikrishnan M Feb 10 '24 at 02:18
  • If the exponent is $\frac 1n$ and $n\to\infty$, why do you consider the limit a $1^\infty$ form? – peterwhy Feb 10 '24 at 02:18
  • Because nothing else is striking me. – Harikrishnan M Feb 10 '24 at 02:19
  • See I don't have any idea of how to do this question other than that. – Harikrishnan M Feb 10 '24 at 02:21
  • Any hints will be appreciated. – Harikrishnan M Feb 10 '24 at 02:21
  • 1
    So you just put in whatever you want, regardless of what your problem is? What you have isn't even an indeterminate of type $1^{\infty}$. It's an indeterminate of type $0^0$. Do you not have any general techniques for solving indeterminate limits of type $0^0$ that rely on actual mathematical process and not silly sayings that clearly do not convey any meaningful information to you? – Arturo Magidin Feb 10 '24 at 02:21
  • But how $0^0$ Some infinity by infinity factorial is there...! – Harikrishnan M Feb 10 '24 at 02:22
  • So what? $(2n)!/n!n^n$ goes to $0$ as $n\to\infty$, for the same reason that $n!/n^n$ goes to $0$ as $n\to\infty$. – Arturo Magidin Feb 10 '24 at 02:23
  • @ArturoMagidin $(2n)!/(n!n^n)$ goes to $\infty$. – peterwhy Feb 10 '24 at 02:46
  • @HarikrishnanM Just check that trick once again from whichever source you got it. "Getting the monkey down" seems to be taking logarithm. But then finally you might have to raise the entire thing to the power of $e$ to get the final answer. You seem to have forgotten that. Reducing $1$ might be something that works in some specific cases, but not all the time. Don't study it as a formula unless you are crystal clear about how and in what specific situations it works. Always understand the logic and derivation behind any math formula. – Sasikuttan Feb 10 '24 at 02:50
  • Ok, I'll check that ! Thank you. – Harikrishnan M Feb 10 '24 at 02:50

1 Answers1

1

Assuming that the limit exists, let's let $$\begin{align} L &= \lim_{n \to \infty} \left(\frac{(2n)!}{n! n^n} \right)^{1/n} \end{align}$$ Then taking the logarithm, $$\begin{align} \ln(L) &= \lim_{n \to \infty} \frac{\ln((2n)!) - \ln(n!) - n\ln(n)}{n} \end{align}$$ I assume this is what you mean by "Getting the monkey down".

The opening wedge here is Stirling's approximation: $$ \ln(n!) = n\ln(n) - n + O(\ln(n))$$ for large $n$. Then asymptotically, $$\begin{align} \frac{\ln((2n)!) - \ln(n!) - n\ln(n)}{n} &= \frac{\left[2n\ln(2n) - 2n \right] - \left[n\ln(n) - n \right] - n\ln(n) + O(\ln(n))}{n}\\ &= \frac{(2\ln(2) - 1)n}{n} + \frac{O(\ln(n))}{n}\\ \Rightarrow\ln(L) = \lim_{n \to \infty}\frac{\ln((2n)!) - \ln(n!) - n\ln(n)}{n} &= 2\ln(2) - 1\\ \Rightarrow L &= e^{2\ln(2) - 1} = \frac{4}{e} \end{align}$$

AlkaKadri
  • 2,130
  • Please don't write answers to duplicate posts. Better to post this in the target duplicate, if an equivalent answer isn't there yet. – Arturo Magidin Feb 10 '24 at 02:26
  • 2
    @ArturoMagidin I agree with this general sentiment, but I guess how much is it incumbent upon me to search for an existing answer in a different post before I try to answer a question? – AlkaKadri Feb 10 '24 at 02:28
  • @AlkaKadri it's plenty incumbent. It's how the site works. – Randall Feb 10 '24 at 02:42
  • 2
    @Randall Searching is something that very specifically doesn't work on this site, so much so that third party tools are necessary. – Theo Bendit Feb 10 '24 at 02:56
  • 1
    I'm not sure I'm a fan of creating an atmosphere where someone that's volunteering their time to answer a question should fear that they've not scoured the site enough for an existing answer. Or that they should be castigated in the event that they've missed one. Clearly there's some consensus/ideology behind how the site should work that I wasn't aware of here. – AlkaKadri Feb 10 '24 at 03:05
  • A link to the duplicate had already been posted when you posted your answer. I realize it probably hadn't when you began writing it, but once you posted and the duplicates identified and linked, you should have moved it there. Now you can't because the OP has accepted it. – Arturo Magidin Feb 10 '24 at 03:09
  • 1
    @ArturoMagidin I did not see this link, but I'll take your word for it. Personally, I would tell you that it feels hardly fair to blame the user of this site for still being able to post an answer after a duplicate has been identified (especially since the I never saw this happen), but I can swallow this criticism more easily than being expected to do research for existing answers. I'll be on my way now. – AlkaKadri Feb 10 '24 at 03:20
  • 1
    I'm not "blaming". Note that I made a request, and explained what a better course of action was. – Arturo Magidin Feb 10 '24 at 03:25