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I have solved the problem in just 2 lines using a theorem which asserts that

"Let ${u_n}$ be a real sequence such that $u_n > 0 \forall n \in \mathbb{N}$ and $\lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \ell$ (finite of infinite). Then $\lim_{n \to \infty} (u_n)^{1 \over n} =\ell$ "

To prove the aforesaid limit, I fix $u_n={n^n \over n!}$. Then $u_n>0 \forall n \in \mathbb{N}$ and $\lim_{n \to \infty}\frac{u_{n+1}}{u_n}= \lim_{n \to \infty}(1+{1 \over n})^n=e$.

Then it follows from the above theorem that $\lim_{n \to \infty} (u_n)^{1 \over n} =e$ i.e. $ \lim_{n \to \infty} \frac{n}{(n!)^\frac{1}{n}} = e $. (Proved)

But I am trying to prove it without using the theorem. I am trying to get a generic proof.

Can anyone provide a proper wayout for this?

Thanks for your help in advance.

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    $\begingroup$ If you assume $\int_0^1\ln(x){\rm~d}x$ exists and equals 1, then you can derive your limit by taking Riemann sums. $\endgroup$ Apr 30, 2018 at 17:16
  • $\begingroup$ With the standard estimate $\ln n! = n\ln n -n + O(\ln n)$ you can easily derive $(n!)^{\frac 1n} = e^{-1}n + o(n)$. $\endgroup$ Apr 30, 2018 at 17:19
  • $\begingroup$ @ SimplyBeautifulArt, can you please explain it more? $\endgroup$
    – MathBS
    Apr 30, 2018 at 17:24
  • $\begingroup$ I took the liberty of changing two occurrences of $u_n+1$ to $u_{n+1}. \qquad$ $\endgroup$ Apr 30, 2018 at 17:27
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    $\begingroup$ @SimplyBeautifulArt, I think you meant the integral equals $-1$, not $1$ (since $\ln x\lt0$ for $0\lt x\lt1$). $\endgroup$ Apr 30, 2018 at 17:31

3 Answers 3

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EDIT: As pointed out in the comments, even though the final inequality is correct, it is insufficient since $(n+1)^{1/n} \to 1$ as $n \to \infty$. The lower bound can be obtained as shown in @adfriedman's answer.

Here's my take on it: Whenever $n \geq 3$, we have $$ n^n \geq (n+1)!, $$ and thus $$ n^n \geq (n+1)n! \quad \Leftrightarrow \quad \frac{n}{n!^{\frac{1}{n}}} \geq (n+1)^{\frac{1}{n}}. $$ On the other hand, the Taylor expansion of $e^n$ gives $$ \frac{n^n}{n!} \leq \sum_{k=0}^{\infty} \frac{n^k}{k!} = e^n \quad \Leftrightarrow \quad \frac{n}{n!^{\frac{1}{n}}} \leq e. $$ So, $$ (n+1)^{\frac{1}{n}} \leq \frac{n}{n!^{\frac{1}{n}}} \leq e. $$ Apply the Squeeze Theorem.

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  • $\begingroup$ Are you sure about $n^n \geq (n+1)!$ ? $\endgroup$ Apr 30, 2018 at 17:44
  • $\begingroup$ I have not proven it formally, but $n^n$ should grow much faster than the factorial, and the inequality seems to start holding for $n=3$. $\endgroup$
    – MSDG
    Apr 30, 2018 at 17:46
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    $\begingroup$ It can be proved by induction, using the binomial expansion. $\endgroup$ Apr 30, 2018 at 17:49
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    $\begingroup$ To show that just use the fact that $(n+1)!=(n+1)\cdots 2 \cdot 1= 2(n+1)[n\cdots 3]$, then $n\cdots 3<n^{n-2}$ and $2(n+1)<n^2$ for $n$ large enough. $\endgroup$ Apr 30, 2018 at 17:50
  • $\begingroup$ Excellent, thanks for completing my arguments! $\endgroup$
    – MSDG
    Apr 30, 2018 at 17:52
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Hint.

Use the Stirling asymptotic approximation formula

$$ n! = \sqrt{2\pi n}\left(\frac{n}{e}\right)^n(1+O(1/n)) $$

because

$$ \frac{n}{(n!)^{1/n}}= \frac{n}{(2\pi n)^{1/(2n)}(n/e)}(1+O(1/n)) = \frac{e}{(2\pi n)^{1/(2n)}}(1+O(1/n)) $$

hence

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

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  • $\begingroup$ Since when does an approximation prove anything about a precise value? $\endgroup$ Apr 30, 2018 at 17:20
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    $\begingroup$ @CesarEo, Sorry, I think it is not a convenient way for solving the problem, as it is a 1st year undergraduate problem, student will be more happy with basic generic proofs. Even as a student I don't know the formula. $\endgroup$
    – MathBS
    Apr 30, 2018 at 17:20
  • $\begingroup$ @CogitoErgoCogitoSum The "approximation" in "Stirlings approximation" is just a name. Stirlings comes with full control over the error and can be applied. The problem with using Strirlings is that it's like assuming the answer: the problem is trivial if we are allowed to use Strirlings. $\endgroup$
    – Winther
    Apr 30, 2018 at 17:24
  • $\begingroup$ @CogitoErgoCogitoSum When it's not an approximation. The correct symbol would be $$\operatorname*{\sim}_{n\to\infty}$$ not $\approx$, and it has a very precise meaning: $$\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n} \xrightarrow[n\to \infty]{}1\,.$$ Sadly, as mentioned above, this is too big of a hammer for this (tantamount to using the hammer to forge the hammer...) $\endgroup$
    – Clement C.
    Apr 30, 2018 at 17:28
  • $\begingroup$ I hear you only need a 1/2 of a lathe to make from it the rest. $\endgroup$ Apr 30, 2018 at 17:59
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First, write $$\ln\left(\frac{n}{(n!)^{1/n}}\right) =\ln(n) - \tfrac{1}{n}\ln(n!) = \ln(n) - \tfrac1n \sum_{k=1}^n \ln(k) = \ln(n) - \tfrac1n S_n$$

Consider upper and lower approximations of the integral of $\ln(x)$:

$$S_n = \sum_{k=1}^n \ln(k) \leq \int_1^{n+1}\ln(x)\;dx \leq \sum_{k=2}^{n+1}\ln(k) = S_n + \ln(n+1)$$ Hence $$S_n \leq \int_1^{n+1}\ln(x)\;dx = \left[ x\ln(x)-x \right]_1^{n+1} = (n+1)\ln(n+1) - n$$ and $$\int_1^{n+1}\ln(x)\;dx - \ln(n+1) = n\ln(n+1) - n \leq S_n$$

It follows: $$\ln(n)-\tfrac1n \big((n+1)\ln(n+1) - n\big) \leq \ln(n) - \tfrac1n S_n \leq \ln(n) - \tfrac1n\big(n\ln(n+1) - n\big) $$ which simplifies to $$\ln\Bigg(\underbrace{\frac{n}{(n+1)^{1+\tfrac1n}}}_{\to \; 1}\Bigg) + 1 \leq \ln(n)- \tfrac1n S_n \leq \ln\Bigg(\underbrace{\frac{n}{n+1}}_{\to\; 1}\Bigg) + 1$$

By the squeeze theorem $\ln\left(\frac{n}{(n!)^{1/n}}\right) = \ln(n) - \tfrac1n S_n \to 1$ as the log terms on both sides approach $\ln(1) = 0$, hence $$\frac{n}{(n!)^{1/n}} = \exp\left(\ln\left(\frac{n}{(n!)^{1/n}}\right)\right) \to \exp(1) = e$$

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