Find $\lim_{n\to \infty}\frac{n}{\sqrt[n]{n!}}$

Question

Find the limit for $\lim_{n\to \infty}\frac{n}{\sqrt[n]{n!}}$

I have no idea how to do this - they always say use the ratio test when there are factorials but the nth root stuff makes it so ugly and i can't get anything to come out of it. Does anyone know what process I should use?

The title is right, you just mention the ratio test, which is not much use for sequences (except that for a sequence to converge to a nonzero value, you must have $x_{n+1}/x_n\to 1$.) — Thomas Andrews, Dec 02 '15 at 16:13
Alternative: https://en.wikipedia.org/wiki/Stirling%27s_approximation — vadim123, Dec 02 '15 at 16:26
@vadim123 Is the equivalent enough here? I haven't worked out the details using Sitrling's, but the fact that a $n$-th root is taken gives me the gut feeling that more terms in the expansion are needed. (?) — Clement C., Dec 02 '15 at 16:36

Kushal Bhuyan · Answer 1 · 2015-12-02T16:40:49.027

2

Result: If $\lim_{n\to \infty}{\frac{a_{n+1}}{a_n}}=l$ then $\lim_{n\to \infty}{\sqrt[n]{a_n}}=l$

Take $(a_n)^{1/n}=\left(\frac{n^n}{n!}\right)^{1/n}$ Now consider $a_n=\frac{n^n}{n!}$ and $a_{n+1}=\frac{(n+1)^{n+1}}{(n+1)!}$.

$$\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\lim_{n\to \infty}\frac{(n+1)^n}{n^n}=\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e$$ hence $$\lim_{n\to \infty}\frac{n}{\sqrt[n]{n!}}=e$$

edited Dec 02 '15 at 16:40

answered Dec 02 '15 at 16:28

Kushal Bhuyan

7,110

Petite Etincelle · Answer 2 · 2015-12-02T16:36:19.003

1

Let $x_n = \frac{n}{\sqrt[n]{n!}}$

$\log(x_n) = \log(n) - \dfrac{\log(n!)}{n} = \dfrac{n\log(n) - \sum_{k=1}^n \log(k)}{n}$

by Stolz–Cesàro theorem, it has the same limit as $(n+1)\log(n+1) - \log(n+1) - n\log(n) = \log\left((1+\dfrac{1}{n})^n\right) \to \log(e) = 1$

edited Dec 02 '15 at 16:36

answered Dec 02 '15 at 16:34

Petite Etincelle

14,767

@ClementC. $\log(x_n)$ conveges to 1, which means $x_n$ converges to $e$ – Petite Etincelle Dec 02 '15 at 16:36
You're right, I had misread what you had done (forgetting the $\log$). – Clement C. Dec 02 '15 at 16:40

score 1 · Answer 3 · answered Dec 02 '15 at 16:37

We have $$ \frac{n}{\sqrt[n]{n!}}=\left(\frac{n^n}{n!}\right)^{1/n}=\exp\left(\frac1n\,\log\left(\frac{n^n}{n!}\right)\right) =\exp\left(\frac1n\,n\log n-\frac1n\,\log n!\right) =\exp\left(\log n-\frac1n\,\sum_{k=2}^n\log k\right). $$ Now we have $$ \int_1^{n-1}\log t\,dt\leq\sum_{k=2}^n\log k\leq\int_1^n\log t\,dt, $$ i.e. $$(n-1)\log n-(n-2)\leq\sum_{k=2}^n\log k\leq n\log n - (n-1).$$ Then $$\log n-\log n+\frac{n-1}n\leq\log n-\frac1n\,\sum_{k=2}^n\log k\leq\log n-\frac{n-1}n\,\log n+\frac{n-2}n,$$ or $$1-\frac{1}n\leq\log n-\frac1n\,\sum_{k=2}^n\log k\leq-\frac{1}n\,\log n+1-\frac2n.$$ Thus $$ \lim_{n\to\infty}\log n-\frac1n\,\sum_{k=2}^n\log k=1, $$ and \begin{align} \lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}&=\lim_{n\to\infty}\exp\left(\log n-\frac1n\,\sum_{k=2}^n\log k\right)\\ &=\exp\left(\lim_{n\to\infty}\log n-\frac1n\,\sum_{k=2}^n\log k\right)\\ &=\exp(1)=e. \end{align}

score 0 · Answer 4 · edited Apr 13 '17 at 12:21

Using an asymptotic expansion of $\ln(n!)$, obtained e.g. as in this post (there, by a comparison series/integral): $$\ln(n!) \operatorname*{=}_{n\to\infty} n\ln n - n +o(n)$$ you get $$ \sqrt[n]{n!} = e^{\frac{1}{n}\ln(n!)} = e^{\frac{1}{n}\sum_{k=1}^n \ln k} = e^{\frac{1}{n}\left( n\ln n - n + o(n)\right)} = e^{\ln n - 1 + o(1)} = n\cdot e^{-1+o(1)} $$ and therefore $$ \frac{n}{\sqrt[n]{n!}} = e^{1+o(1)} \xrightarrow[n\to\infty]{} e. $$

Find $\lim_{n\to \infty}\frac{n}{\sqrt[n]{n!}}$

4 Answers4