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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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3 Answers 3

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

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