How to see that $\lim \frac{k}{\sqrt[k]{k!}}=e$?

Question

There is a way to see that $\lim \frac{k}{\sqrt[k]{k!}}=e$ from these definitions?

$$e=\lim\left(1+\frac1k\right)^{\frac1k};\quad e=\sum\frac1{k!}$$

If it is not possible, how to deal with the limit of the title to get it limit? Thank you.

Related: http://math.stackexchange.com/questions/28476/finding-the-limit-of-frac-n-sqrtnn — Henry, Oct 31 '16 at 11:24

score 2 · Answer 1 · edited Apr 13 '17 at 12:19

2

We can use the fact that if $a_n>0$ for all $n\ge1$ and the sequence $\frac{a_{n+1}}{a_n}$ converges in $[0,\infty]$, then $$ \lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n} $$ (see this answer).

We have that $$ \frac n{\sqrt[n]{n!}}=\biggl(\frac{n^n}{n!}\biggr)^{1/n} $$ and let us denote $a_n=n^n/n!$. Then $$ \frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}n!}{(n+1)!n^n}=\frac{(n+1)^n}{n^n}=\biggl(1+\frac1n\biggr)^n\to e\quad\text{as}\quad n\to\infty. $$ Hence, $$ a_n^{1/n}=\frac n{\sqrt[n]{n!}}\to e\quad\text{as}\quad n\to\infty. $$

edited Apr 13 '17 at 12:19

Community

1

answered Oct 31 '16 at 11:19

Cm7F7Bb

17,364

Yes, this is the origin of my question, the equivalence in the calculus of the radius of convergence of a power series when $a_k=\frac{k!}{k^k}$. This is why I want to know if we can evaluate the limit from a different result. – Masacroso Oct 31 '16 at 11:20

E. Joseph · Accepted Answer · 2016-10-31T11:17:46.157

1

Hint

(it does not follow directly from your definitions though)

You can use Stirling formula:

$$k!\underset{k\to\infty}{\sim} \sqrt{2\pi k}\frac{k^k}{e^k}.$$

edited Oct 31 '16 at 11:17

answered Oct 31 '16 at 11:14

E. Joseph

14,843

I dont know the topic of Stirling approximation very well, I assume that $\sim$ means equivalence in the infinity, right? – Masacroso Oct 31 '16 at 11:16
@Masacroso Yes that right. (it means approximately that the quotient tends to $1$) – E. Joseph Oct 31 '16 at 11:17
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@Masacroso It means that the ratio of the both sides tends to $1$ (not approximately, but exactly). – Cm7F7Bb Oct 31 '16 at 11:23
@Cm7F7Bb It is only true if one of the sides is never equal to $0$. That's why I said approximately. – E. Joseph Oct 31 '16 at 11:24

score 1 · Answer 3 · answered Oct 31 '16 at 14:42

1

Using $$ \ln\frac{k}{\sqrt[k]{k!}}=\ln k-\frac1k\ln(k!)=-\sum_{i=1}^{k}\frac{1}{k}\ln\frac{i}{k}\to-\int_0^1\ln xdx=1$$ as $k\to\infty$, one has $$ \lim_{k\to\infty}\frac{k}{\sqrt[k]{k!}}=e. $$

answered Oct 31 '16 at 14:42

xpaul

44,000

How to see that $\lim \frac{k}{\sqrt[k]{k!}}=e$?

3 Answers3