The denominator dominates the numerator so the term $\displaystyle \frac {n!}{n^n}$ definitely tends to $0$, and so does the power $\displaystyle\frac {1}{n}$.

So, overall it is $\displaystyle0 ^ 0$ form.

I proceeded by taking its $log$ and then I got stuck because I found it difficult to solve by L' Hopital's rule.

  • $\begingroup$ This is difficult for me to follow. :) $\endgroup$
    – Gaurav
    Dec 10, 2014 at 16:44
  • $\begingroup$ Oops... Sorry. I meant to link this. $\endgroup$ Dec 10, 2014 at 16:58

3 Answers 3


Hint: $$n!\approx \left(\dfrac{n}{e}\right)^n\cdot\sqrt{2n\pi},n\to \infty$$

hint 2

$$\left(\dfrac{n!}{n^n}\right)^{\frac{1}{n}}=e^{\sum_{i=1}^{n}\dfrac{1}{n}\ln{\dfrac{i}{n}}}$$ so $$\lim_{n\to\infty}\sum_{i=1}^{n}\dfrac{1}{n}\ln{\dfrac{i}{n}}=\int_{0}^{1}\ln{x}dx$$ hint 3:

$$\lim_{n\to\infty}\sqrt[n]{a_{n}}=\lim_{n\to \infty}\dfrac{a_{n+1}}{a_{n}}$$ let $$a_{n}=\dfrac{n!}{n^n}$$


Using L'Hospital's rule for this is invalid, because $n!$ is an integer-only function without a derivative.

You're idea of using logarithms is a good one, though.

$\lim\limits_{n\to\infty} \frac{n!}{n^n}^\frac{1}{n}=e^{\lim\limits_{n\to\infty} ln\frac{n!}{n^n}^\frac{1}{n}}=e^{\lim\limits_{n\to\infty} \frac{1}{n}\sum\limits_{i=1}^n\ln i-\ln n}$

$\ln(x)$ is a monotone strictly increasing function, so $\sum\limits_{i=1}^n\ln i\leq\int\limits_1^{n+1}ln(x)dx$. You can verify this graphically. Playing around with the integrals a bit, we can also get a lower bound of $\int\limits_1^n\ln(x)dx$. You can also verify this. Then:

$n\ln n-n+1\leq\sum\limits_{i=1}^n\ln i\leq(n+1)\ln(n+1)-n$.

Dividing by $n$ and subtracting $\ln n$ yields:

$-1+\frac{1}{n}\leq\frac{\sum\limits_{i=1}^n\ln i}{n}-\ln n\leq\frac{n+1}{n}\ln(n+1)-1-\ln n=\frac{n+1}{n}(\ln(n+1)-\ln n)+\frac{n+1}{n}(\ln n)-1-\ln n=\frac{n+1}{n}(\ln(n+1)-\ln n)+\frac{1}{n}(\ln n)-1$

We can now get an inequality about limits from this expression.

$-1\leq\lim\limits_{n\to\infty}\frac{\sum\limits_{i=1}^n\ln i}{n}-\ln n\leq1\times0+0-1=-1$

We can see that $\lim\limits_{n\to\infty}\frac{\sum\limits_{i=1}^n\ln i}{n}-\ln n=-1$, which finally lets us solve the original limit.

$\lim\limits_{n\to\infty} \frac{n!}{n^n}^\frac{1}{n}=e^{-1}=\frac{1}{e}$


Using Stloz-Cesaro: $$\eqalign{\log L & = \lim_{n\to\infty}\log{\frac{n!}{n^n}}^\frac{1}{n} = \lim_{n\to\infty}\frac{\log 1+\log 2+\cdots+\log n-n\log n}n=\cr & = \lim_{n\to\infty}(n+1)\log(n+1)-n\log n+\log(n+1)=\cdots }$$


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