MHB Limit of (n)^(1/n)/n as n approaches infinity

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The discussion focuses on determining the limit of (n!)^(1/n)/n as n approaches infinity. Participants share various methods to solve the problem, with one user expressing appreciation for another's solution. Hints are provided for those who prefer not to use Stirling's approximation. The conversation highlights a collaborative effort to explore mathematical solutions. Overall, the thread emphasizes problem-solving techniques in calculus.
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Determine $$\lim_{{n}\to{\infty}}\frac{(n!)^{1/n}}{n}$$
 
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Rido12 said:
Determine $$\lim_{{n}\to{\infty}}\frac{(n!)^{1/n}}{n}$$

By the continuity of the natural logarithm and using Stirling's approximation, we find that:
$$\ln \left [ \lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n} \right ] = \lim_{n \to \infty} \ln \frac{(n!)^{\frac{1}{n}}}{n} = \lim_{n \to \infty} \left [ \frac{\ln{n!}}{n} - \ln{n} \right ] = \lim_{n \to \infty} \left [ \left ( \ln{n} + O \left ( \frac{\ln{n}}{n} \right ) - 1 \right ) - \ln{n} \right ] = -1$$
Therefore it follows that:
$$\lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n} = e^{-1} = \frac{1}{e}$$
 
Excellent solution Bacterius! Thanks for participating. :D

For anyone who wants to tackle this without Stirling's approximation, I have a few hints:

Use the fact that, assuming $f$ is non-negative and increasing:
$$\sum_{k=1}^{n-1}f(k)\le\int_1^nf(x) \,dx\le\sum_{k=1}^{n}f(k)$$

and apply it to $\ln(x)$ to get $en^ne^{-n}<n!<en^{n+1}e^{-n}$.
 
Here is my method:

If $a_n = \frac{(n!)^{1/n}}{n}$, then

$$a_n = \sqrt[n]{\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\cdots \left(1 - \frac{n-1}{n}\right)} = \exp\left\{\frac{1}{n}\sum_{k = 0}^{n-1}\log\left(1 - \frac{k}{n}\right)\right\},$$

which converges to

$$\exp\left\{\int_0^1 \log(1 - x)\, dx\right\} = \exp\left\{\int_0^1\log x\, dx\right\} = \exp(-1) = \frac{1}{e}.$$
 
Hi Euge! Very interesting solution, excellent job! (Cool)

$$\sum_{k}^{n-1}\ln k \le\int_{1}^{n} \ln x\,dx \le \sum_{k=1}^{n}\ln k$$
$$\implies \ln(n-1)! \le n \ln n-n+1 \le \ln n!$$

On the left hand side, it can be shown that $\ln(n-1)! \le n\ln n -n+1 \implies n! \le n^{n+1}e^{-n}e$ and on the right hand side, $n \ln n-n+1 \le \ln n! \implies n^ne^{-n}\le n!$.

$$n^ne^{-n}e \le n! \le n^{n+1}e^{-n}e$$$$\implies\frac{e^{1/n}}{e}\le \frac{(n!)^{1/n}}{n} \le \frac{e^{1/n}n^{1/n}}{e}$$

Furthermore, $\lim_{{n}\to{\infty}}a^{1/n}=1$ and $\lim_{{n}\to{\infty}}n^{1/n}=1$, so by the squeeze theorem,
$$\lim_{{n}\to{\infty}}\frac{(n!)^{1/n}}{n}=\frac{1}{e}$$
 
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