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I am trying to prove that $\frac{n!}{n^n}$ converges to $0$.

I write the sequence as $$\frac{n!}{n^n} = \frac{1 \cdot 2 \cdot 3 \ldots n }{n \cdot n \cdot n \ldots n}$$ Next I break the expression into $$(\frac{1}{n}) \cdot (\frac{2}{n}) \cdot (\frac{3}{n}) \ldots (\frac{n-1}{n}) \cdot (\frac{n}{n})$$ Here comes my issue: the terms when taken from the beginning are $(\frac{1}{n})$, $(\frac{2}{n})$ and so on all converge to $0$. But when taken from the end the terms $(\frac{n}{n})$, $(\frac{n-1}{n})$, $(\frac{n-2}{n})$ and so on, all converge to $1$.

How is that possible?

FD_bfa
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Adam
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    Why is that a problem? So far this is not a contradiction with the statement you want to prove. – Michal Adamaszek Dec 07 '22 at 10:31
  • Well, when using combination rules for convergent sequences (product rule), I either get that the sequence converges to 0 or 1. – Adam Dec 07 '22 at 10:35
  • I am a little confused... Either all the factors converge to 0 or they all converge to 1, right? Which is it? – Adam Dec 07 '22 at 10:48
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    Your argument shows $\frac{n!}{n^n}\le\frac{1}{n}$ because all the other factors are positive and not larger than $1$. Hence the limit must be $0$. The first terms do not converge to $1$ , this is the fallacy. – Peter Dec 07 '22 at 10:49
  • I see. 123456789 – Adam Dec 07 '22 at 10:51
  • For any fixed k, $(\frac{n-k}{n})$ indeed converge to 1. But that's not what you have in your expression because k is not fixed. See Peter comment for a solution to the convergence problem – F.Carette Dec 07 '22 at 10:51
  • The product rule does not apply here because you do not have a product of a finite number of sequences, but a longer and longer product. – Michal Adamaszek Dec 07 '22 at 11:05
  • Related (not quite a dupe): https://math.stackexchange.com/questions/2520880/proof-that-lim-limits-n-to-infty-fracnnn-is-0/ – Theo Bendit Dec 07 '22 at 11:06
  • One last thing I dont understand: the initial terms converge to 0 and the final terms converge to 1. Where is the exact place where this change happens? – Adam Dec 07 '22 at 11:10
  • The final terms converge to $1$, but the product of the last few terms does not. The product of the last two terms is $(n^2-n)/n^2$, and of the last three is $(n^3-3n^2+n)/n^3$, and so forth. All of those products are less than $n$, assuming $n$ is a positive integer. – Eric Snyder Dec 07 '22 at 11:27
  • When you say "final terms" you forget this is the $n$-th term, where $n$ grows. For every fixed $k$ the $k$-th term converges to $0$. (For example the 33-th term $33/n$ converges to $0$) But when you say the "last term" you really mean the $n$-th term and that is a different type of subsequence. – Michal Adamaszek Dec 07 '22 at 11:56
  • "Where is the exact place where this change happens" That is like the question of at what point does removing a grain of sand make a pile of sand no longer a sandpile. The easiest answer is "don't worry about it." It happens "somewhere in the middle" but it really doesn't help to pin down exactly. Similarly here, if we look at "the middle" of your terms, you have $\frac{n/2}{n}$ and the terms around it converge to $\frac{1}{2}$... There is a whole range of values here. – JMoravitz Dec 07 '22 at 13:15
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    If you look at values near the $0%$ mark of your range, you'll find they converge to $0$. If you look at values near the $100%$ mark of your range, you'll find they converge to $1$. If you look at values near the $50%$ mark of your range, you'll find they converge to $0.5$. If you look at values near the $x%$ mark of your range you'll find they converge to $\frac{x}{100}$ and so on... It is not so much that it "shifts from 0 to 1" at a specific place, it changes gradually across the whole interval. – JMoravitz Dec 07 '22 at 13:17
  • I'm guessing this is either forgotten or satisfactorily answered but since the question has gotten bumped, the problem is perhaps speaking of the terms at all, as if there is a definite set of sequences whose termwise product is $n!/n^n$. – Andrew D. Hwang May 11 '24 at 23:22
  • So many comments but the important and straightforward fact is $\displaystyle {n! \over n^{n}} \stackrel{{\rm as}\ n \to \infty}{\sim} \sqrt{2\pi}\ n^{1/2}{\rm e}^{-n}$ – Felix Marin May 12 '24 at 01:01

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