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Find the limit :

$\lim_{ n \to \infty} (\frac{n!}{n^n})^{\frac{1}{n}}$

My working :

Let $$t = \lim_{ n \to \infty} (\frac{n!}{n^n})^{\frac{1}{n}}$$

Now taking log on both sides :

$$\log t = \frac{1}{n}\log \frac{n!}{n^n}$$ ( we will consider the limit later on )

$$\log t = \frac{1}{n}( \log n! - \log n^n)$$

$$\log t = \frac{1}{n}\log n! -\frac{n}{n} \log n $$

$$\log t = \lim_{ n \to \infty} \frac{1}{n}\log n! - \lim_{ n \to \infty} \log n $$

$$\log t = 0 - \lim_{ n \to \infty} \log n $$

I know its duplicate problem at this site but I want to do it in more easier way by taking logs rather using Cauchy-d'Alembert criterion, Stolz–Cesàro theorem, or The Lalescu sequence

Can you please help me from here... thanks...

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  • $\begingroup$ You shouldn't write $\log t$ before showing the existence of $t$. $\endgroup$ – Philippe Malot Oct 25 '13 at 17:02
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STolz theorem

$$\lim_{n\to\infty} \log t=\lim_{n\to\infty} (\log (n+1)-(n+1)\log (n+1)+n\log n)=-\lim_{n\to\infty} \log (1+\frac1n)^n=-1$$

so

$$\lim_{ n \to \infty} (\frac{n!}{n^n})^{\frac{1}{n}}=\frac 1e$$

another method

Lemma if $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=l$, then $\lim\limits_{n\to\infty}\sqrt[n]{a_n}=l$

Let $a_n=\frac{n^n}{n!}$, it is easy to see

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}(1+\frac1n)^n=e$$

so

$$\lim_{n\to\infty}\sqrt[n]{a_n}=e$$

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  • $\begingroup$ Your second method is the clearest way, I think! It gives a good intuition for why this is true. $\endgroup$ – Peter LeFanu Lumsdaine Oct 26 '13 at 20:29
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See Stirling's approximation.

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Using inequality $$ (1+\frac{1}{n})^n < e$$ is shown by mathematical induction inequality $$(\frac{n}{e})^n <n!<e(\frac{n+1}{e})^{n+1}$$ where $n$ is a integer positive strictly. From this inequality it follows that $$\frac{1}{e} <(\frac{n!}{n^n})^{\frac{1}{n}}<\frac{1}{e}.\frac{n+1}{n}.(n+1)^\frac{1}{n}.$$Now applying Sandwich Theorem we obtain $$\lim (\frac{n!}{n^n})^{\frac{1}{n}} =\frac{1}{e}.$$ (It is a solution that uses the simplest means)

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Let $$L=\lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$ and consider $\ln(L)$. $$\ln(L)=\lim_{n\rightarrow\infty}\frac{1}{n}\ln\left(\frac{n!}{n^n}\right)\\=\lim_{n\rightarrow\infty}\frac{1}{n}\left[\ln\left(\frac{1}{n}\right)+\ln\left(\frac{2}{n}\right)+\dots+\ln\left(\frac{n-1}{n}\right)+\ln\left(\frac{n}{n}\right)\right]\tag{1}$$ The expression $(1)$ is the Riemann sum equal to the integral $$\int_0^1\ln(x)dx=\ln(L)\tag{2}$$ The integral $(2)$ is equal to $$\left[x\ln(x)-x\right]^1_0=\lim_{x\rightarrow 0^+}-x\ln(x)-1\tag{3}$$ To evaluate the limit $(3)$, we can use l'Hôpital $$\lim_{x\rightarrow 0^+}\frac{\ln(x)}{\frac{1}{x}}=\lim_{x\rightarrow0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\rightarrow 0^+}-x=0$$ Therefore $\ln(L)=0-1=-1$, and so it follows that $L=e^{-1}$.

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