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I have a problem figuring out the limit: $\lim_{n \rightarrow \infty}\sqrt[n]{n!}$.

And other similar limits such as:

  1. $\lim_{n \rightarrow \infty} \frac{n}{\sqrt[n]{n!}}$

  2. $\lim_{n \rightarrow \infty}\sqrt[n]{kn \choose n}$

  3. $\lim_{n \rightarrow \infty}\frac{n!}{(n+1)(n+1)...(2n)}$

I also have difficulties proving that if $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}=l>1$, then, $\lim_{n \rightarrow \infty}a_n = \infty$

Can anyone help? Thanj you, Tali

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5 Answers 5

Problems involving nth roots of n! can be solved easily using Stirling's approximation.

Stirling proved that

$$ \lim_{n \to \infty} \frac{e^{-n}n^n\sqrt{2n\pi}}{n!} = 1 $$

Therefore in a limit to infinity, you can replace $n!$ with the numerator in the above expression $\left(e^{-n}n^n\sqrt{2n\pi}\right)$ and the find the nth root.

So, $$ \lim_{n \to \infty} \frac{n}{\sqrt[n]{n!}} = \lim_{n \to \infty} \frac{ne}{{n}} = e $$

Make sure that you see how $\sqrt{2n\pi}$ contributes only a $1$ to the limit. It's easy.

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Thank you very much! – user106445 Nov 9 '13 at 12:06

Use cauchys theorem on the limit

$y=(n!)^{1\over n},, \log y={\log1+\log2+\log n\over n}$

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In responds to the line "I have a problem figuring out the limit: $\lim_{n \rightarrow \infty}\sqrt[n]{n!}$. "

Another way to approach the problem:

$ (n!)^2=\{1 \cdot2 \cdot \cdot \cdot \cdot (n-2)(n-1)n\}\{n(n-1)(n-2) \cdot \cdot \cdot \cdot 3\cdot 2 \cdot 1\}=\{(1 \cdot n)(2 \cdot \overline{n-1})(3 \cdot \overline{n-2}) \cdot \cdot \cdot (n \cdot 1)\}\\ \ge n \cdot n \cdot n \cdot \cdot \cdot n (\text{n times})\\=n^n \\\implies (n!)^{1/n} \ge n^{\frac 12 }\to \infty \,\,\text{as}\,\, n \to \infty $

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Thank you very much! – user106445 Nov 9 '13 at 12:06
@user106445 you are welcome. – learner Nov 9 '13 at 12:39
Nice solution learner..... – juantheron Nov 9 '13 at 13:32

I will use one result to prove this which is the following: if $\{a_n\}$ converges to $l$ ,then $S_n=\frac {a_1+a_2+...a_n}{n}$ also converges to $l$. From which we get that $$\frac {1+1/2+...1/n}{n}$$ converges to $0$. Now using A.M $\geq$ G.M we get $$\frac{1+1/2+...+1/n}{n}\geq \frac {1}{n!^{1/n}}$$ which implies that $$ {n!}^{1/n} \geq \frac{n}{1+1/2+...1/n}$$ But the RHS goes to $\infty$. Hence so is the LHS.

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For 1. note the reciprocal is $(n!/n^n)^{1/n}.$ Apply $\ln $ to that to get

$$\frac{1}{n}\sum_{k=0}^{n-1}\ln (1-k/n) \to \int_0^1\ln (1-x)\, dx = -1.$$

Thus the reciprocal $\to e^{-1},$ which gives $e$ for the limit of the original expression.

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