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How to calculate $\lim_{n\to \infty } \frac{n^n}{n!^2}$?

I tried to solve this by using Stirling's formula but I got stuck at $\lim\limits_{n\to \infty } \frac{e^{2n}}{n^{n+1}}$. Any help?

Vivek
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4 Answers4

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Most of the time you do not need the exact Stirling approximation, just use the simplest:

Factorial Inequality problem $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$

It ensues that $\dfrac{(n!)^2}{n^n}>\left(\dfrac n9\right)^n\to\infty$ so the inverse is going to $0$.

Anyway regarding the point where you are stuck $\dfrac{n^{n+1}}{e^{2n}}=\underbrace{n}_{\to\infty}\bigg(\ \underbrace{\dfrac n{e^2}}_{\to\infty}\ \bigg)^{\underbrace{n}_{\to\infty}}\to \infty$

so the inverse is going to $0$.

zwim
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  • Thanks, one more question, the exact Stirling's approximation will not work in this question? – Vivek Jan 11 '18 at 01:52
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    It would work, this is just more calculations to do and possible mistakes on the way. Just say $n!\sim\sqrt{2\pi n}\ n^ne^{-n}$ then $n!>\left(\frac ne\right)^n$ and conclude exactly like i did with $3$ in place of $e$. – zwim Jan 11 '18 at 01:52
  • why $n!>\left(\frac ne\right)^n$ is Stirling's formula is less than $n!$ ? – Vivek Jan 11 '18 at 01:58
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    $\sqrt{2\pi n}>1$ – zwim Jan 11 '18 at 01:59
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    @Vivek I added details about the point where you are stuck. Is it clear for you now ? – zwim Jan 11 '18 at 02:12
  • yes... Ohh thank you so much – Vivek Jan 11 '18 at 02:15
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Alternative approach. Let $a_n=\frac{n^n} {n! ^2} $. It is clear that $a_n$ is nonnegative. Moreover we have \begin{align} \frac{a_{n+1} } {a_n} &=\frac{(n+1)^{n+1}n!^2}{n^n(n+1)!^2}\\ &=\frac{1}{(n+1)}\left(1+\frac{1}{n}\right)^n\to 0\end{align} Hence $\sum_{n\geq 1} a_n$ converges by the ratio test and that implies $\lim_{n\to \infty} a_n=0$.

Shashi
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Go on! $$ \lim_{n\to\infty}\frac{e^{2n}}{n^{n+1}}=\lim_{n\to\infty}\frac{1}{n}\left(\frac{e^2}{n}\right)^n $$ For $n>18$ you have $$ \frac{e^2}{n}<\frac{1}{2} $$

Alternatively, consider the ratio $$ \frac{(n+1)^{n+1}/((n+1)!)^2}{n^n/(n!)^2}= \left(\frac{n!}{(n+1)!}\right)^{\!2}\left(\frac{n+1}{n}\right)^{\!n}(n+1) =\frac{1}{n+1}\left(1+\frac{1}{n}\right)^{\!n}\to0 $$ Therefore the series $$ \sum_{n\ge1}\frac{n^n}{(n!)^2} $$ is convergent by the ratio test, in particular its general term has limit $0$.

egreg
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Note that

$$a_n=\frac{n^n}{n!^2}=\frac{n^n}{(n+1)^n}\frac{(n+1)!^2}{n!^2(n+1)}\frac{(n+1)^{n+1}}{(n+1)!^2}=\frac{n+1}{\left({1+\frac1n}\right)^n}\frac{(n+1)^{n+1}}{(n+1)!^2}\geq\frac{n+1}{e}a_{n+1}>\frac{n}{e}a_{n+1}$$

thus $$a_{n+1}<\frac{e}{n}a_n$$

and $$a_1=1\implies a_2<\frac{e}{1}\implies a_3<\frac{e}{2}\frac{e}{1}\implies...\implies a_{n+1}<\frac{e^n}{n!}\to0 $$

thus for squeeze theorem

$$0\leq a_{n+1} <\frac{e^n}{n!}\implies \lim\limits_{n\to \infty } \frac{e^{2n}}{n^{n+1}}=0$$

user
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