Finding $\lim_{n\to\infty}\frac{2^n\cdot n!}{n^n}$

Question

$\lim_{n\to\infty}\frac{2^n\cdot n!}{n^n}$

I think we can do it by squeeze rule but I could not find a sequence bigger than this one converging to 0 without using Stirling approximation of n!

score 8 · Answer 1 · answered Nov 20 '17 at 06:04

8

Being a sequence of positive terms you can use D'Alambert's criterion. $$\frac{a_{n+1}}{a_n}=\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}\frac{n^n}{2^n n!}=2\left(\frac{n}{n+1}\right)^n\longrightarrow 2e^{-1}<1$$ and so $a_n=\frac{2^n n!}{n^n}\longrightarrow0$.

answered Nov 20 '17 at 06:04

Alejandro Nasif Salum

5,703

Is the ratio test the same thing as d'alemberts criterion? – Student number x Nov 20 '17 at 06:06
Yes, it is. D'Alambert's quotient/ratio test/criterion. – Alejandro Nasif Salum Nov 20 '17 at 06:14

score 4 · Answer 2 · answered Nov 20 '17 at 06:42

4

$$ \begin{align} \lim_{n\to\infty}\frac1n\log\left(\frac{2^nn!}{n^n}\right) &=\log(2)+\lim_{n\to\infty}\frac1n\sum_{k=1}^n\log\left(\frac kn\right)\\ &=\log(2)+\int_0^1\log(x)\,\mathrm{d}x\\[9pt] &=\log(2)-1 \end{align} $$ Therefore, $$ \lim_{n\to\infty}\left(\frac{2^nn!}{n^n}\right)^{1/n}=\frac2e $$ and so $$ \lim_{n\to\infty}\left(\frac{2^nn!}{n^n}\right)=0 $$

answered Nov 20 '17 at 06:42

robjohn

345,667

I dig this $\qquad$ – user217285 Nov 20 '17 at 07:37

score 3 · Answer 3 · answered Nov 20 '17 at 06:15

3

$$\dfrac{1+2+\cdots+n}{n}\geq\sqrt[n]{n!}$$ then $$\sqrt[n]{\dfrac{2^nn!}{n^n}}=\dfrac{2}{n}\sqrt[n]{n!}<\dfrac{n+1}{n}\to1$$ as $n\to\infty$. This shows the series $\displaystyle\sum{\dfrac{2^nn!}{n^n}}$ is converge and then $\displaystyle\lim_{n\to\infty}{\dfrac{2^nn!}{n^n}}=0$.

answered Nov 20 '17 at 06:15

Nosrati

29,995

1

this looks like a case where $n$-th root test is indeterminate to me – Rolf Hoyer Nov 20 '17 at 06:23
1

Actually, nothing's been proved. Just that if the limit of the n-th root of the general term exists it is least or equal to one. If it were equal to 1, then the root test would be inconclusive indeed. – Alejandro Nasif Salum Nov 20 '17 at 06:26
Not equal, The inequality is strickly for large $n$. – Nosrati Nov 20 '17 at 06:27
The limit is $2e^{-1}$, by the way, since it has to be equal to the limit of the ratio. – Alejandro Nasif Salum Nov 20 '17 at 06:27
But $a_n<b_n$ does not imply $\lim a_n < \lim b_n$. Let, for instance $a_n=1+\tfrac1n$ and $b_n=1+\tfrac2n$. – Alejandro Nasif Salum Nov 20 '17 at 06:29

score 1 · Answer 4 · answered Nov 20 '17 at 06:08

We observe:

$\sum_{n=1}^\infty \frac{2^{n}\cdot n!}{n^n}$

Since, $\frac{2^{n+1}(n+1)!\cdot n^n}{(n+1)^{n+1}\cdot 2^n\cdot n!}=2\cdot (\frac{n}{n+1})^n=2(1-\frac{1}{n+1})^n\to 2e^{-1}<1$ (Using the ratio test) we conclude that the series converges. Therefore:

$\lim_{n\to\infty} \frac{2^n\cdot n!}{n^n}=0$

score 0 · Answer 5 · answered Nov 20 '17 at 05:57

0

Hint: use Stirling's approximation for $n!$

answered Nov 20 '17 at 05:57

Ross Millikan

374,822

score 0 · Answer 6 · answered Nov 20 '17 at 06:37

I have a nontraditional answer that avoids any knowledge of $(1+1/n)^n$ (at least directly).

Note that for each $i = 1, \ldots, n$ there is a unique integer $1\le j \le 6$ such that $(j-1)n/6 < i \le jn/6$. For any given $j$ there are at most $k = \lceil n/6 \rceil$ corresponding $i$. If we bound each $i$ from above by the corresponding multiple of $n/6$, this gives us: $$ n! \le \left(\frac{n}{6}\right)^n2^k3^k4^k5^k6^k = \left(\frac{n}{6}\right)^n (6!)^k $$ Now we have $$\frac{2^nn!}{n^n} \le \frac{2^n}{n^n} \left(\frac{n}{6}\right)^n (6!)^k = \frac{(6!)^k}{3^n} = 3^{6k-n}\left(\frac{(6!)}{3^6}\right)^k \le 3^5\left(\frac{720}{729}\right)^k\,.$$ Here we use that $6k-n = 6\lceil n/6\rceil - n \le 5$.

Note that this converges to 0 as $n\to \infty$, as $n\to \infty$ implies $k \to \infty$.

Finding $\lim_{n\to\infty}\frac{2^n\cdot n!}{n^n}$

6 Answers6