Show that: $$\left(\dfrac{n}{n+1}\right)^{n+1}<\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}<\left(\dfrac{n}{n+1}\right)^n$$ where $n\in \Bbb N^{+}.$

If this inequality can be proved, then we have $$\lim_{n\to\infty}\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}=\dfrac{1}{e}.$$

But I can't prove this inequality. Thank you.

  • $\begingroup$ Have you tried induction? $\endgroup$ – user85798 Jan 20 '14 at 4:25
  • $\begingroup$ yes,I have try it,at last I failed $\endgroup$ – china math Jan 20 '14 at 4:27
  • $\begingroup$ I was able to prove the first inequality only for large enough $n$, and the limit formula together. But, the second inequality is extremely tight, and I do not know how to get it. $\endgroup$ – Sungjin Kim Jan 20 '14 at 23:33
  • $\begingroup$ Finally got both inequalities for large enough $n$. The order of magnitude of error in the first inequality is $O(1/n)$, and the second inequality with $O((\log^2(n))/n^2)$. $\endgroup$ – Sungjin Kim Jan 21 '14 at 5:44
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    $\begingroup$ @nanchangjian You say in your bounty offer that we can use the AM-GM inequality to solve this problem...Can you explain why you're sure that we can find a solution using only the inequality? $\endgroup$ – user37238 Jan 29 '14 at 9:26


Applying Stolz–Cesàro theorem

$$L=\lim_{n\to \infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n\to \infty} \frac{\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}}{(n+1)-n}=\lim_{n\to \infty} \left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)$$

We have $$\ln L=\lim_{n\to \infty} \left(\frac{1}{n}\sum_{i=0}^{n}\ln\left(\frac{i}{n}\right)\right)=\int_{0}^{1}\ln x\,dx=-1$$

$$\to L=\frac{1}{e}$$

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    $\begingroup$ But,we can't prove this limit $\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}$ is exsit $\endgroup$ – china math Jan 20 '14 at 4:51
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    $\begingroup$ and you not anwser my problem,My problem is inequality,But thank you all the same $\endgroup$ – china math Jan 20 '14 at 4:52
  • $\begingroup$ @chinamath: the limit exists by squeeze lemma and is equal to $e^{-1}$, but unfortunately this doesn't quite answer your question. $\endgroup$ – Alex Jan 20 '14 at 5:05
  • $\begingroup$ @Alex: squeezed between which functions? $\endgroup$ – robjohn Jan 28 '14 at 8:44
  • $\begingroup$ @robjohn: I think Alex means that existence of the limit is guaranteed by the inequalities. But, Mrnhan's answer lacks the proof. $\endgroup$ – Sungjin Kim Jan 29 '14 at 19:46

Consider $$ \lim_{n\to\infty}\frac{\log(n!)-n\log(n)}{n}\tag{1} $$ Using Stolz-Cesàro, this is $$ \begin{align} &\lim_{n\to\infty}\frac{\big[\log((n+1)!)-(n+1)\log(n+1)\big]-\big[\log(n!)-n\log(n)\big]}{[n+1]-n}\\ &=\lim_{n\to\infty}n\log\left(\frac{n}{n+1}\right)\\[6pt] &=-1\tag{2} \end{align} $$ Therefore, $$ \lim_{n\to\infty}\frac{n!^{\frac1n}}{n}=\frac1e\tag{3} $$ Inverting $\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n=e^x$, we have $$ \lim_{n\to\infty}n\left(x^{1/n}-1\right)=\log(x)\tag{4} $$ Using the equation $$ \begin{align} (n+1)!^{\frac1{n+1}}-n!^{\frac1n} &=\frac{n!^{\frac1n}}{n+1}(n+1)\left(\left(\frac{n+1}{n!^{\frac1n}}\right)^{\frac1{n+1}}-1\right)\tag{5} \end{align} $$ and $(3)$ and $(4)$, we get $$ \begin{align} \lim_{n\to\infty}(n+1)!^{\frac1{n+1}}-n!^{\frac1n} &=\lim_{n\to\infty}\frac{n!^{\frac1n}}{n+1}(n+1)\left(\left(\frac{n+1}{n!^{\frac1n}}\right)^{\frac1{n+1}}-1\right)\\ &=\lim_{n\to\infty}\frac1e(n+1)\left(e^\frac1{n+1}-1\right)\\[9pt] &=\frac1e\log(e)\\[9pt] &=\frac1e\tag{6} \end{align} $$ I'm still working on a simple derivation of the initial inequality, however.

  • $\begingroup$ I think that the second line of (6) need to be justified in detail, since you replaced only some terms involving $n$ by the limits. So, there must be error estimate when you do that. $\endgroup$ – Sungjin Kim Jan 27 '14 at 2:37
  • $\begingroup$ @i707107: The sequence of functions $f_n(x)=n\left(x^{1/n}-1\right)$ converge equicontinuously for all $x$, in particular at $x=e$. Thus, $$\lim_{n\to\infty}f_n(x_n)=\log\left(\lim_{n\to\infty}x_n\right)=\log(e)$$ $\endgroup$ – robjohn Jan 27 '14 at 3:30
  • $\begingroup$ @robjohn Try to read a starting problem better. I think you did not answer the question. -1. $\endgroup$ – Michael Rozenberg Jan 18 '17 at 5:02
  • $\begingroup$ @MichaelRozenberg: if you read the answers that were given 2 years ago, no one was able to answer the initial question, so they answered the question that the OP was trying to answer with the initial inequality. As you see if you read my answer I note that this is not a full answer, and I stated that I was still looking for a proof of the initial inequality. The bounty is working; I have been trying to show the initial inequality again. $\endgroup$ – robjohn Jan 18 '17 at 10:07
  • $\begingroup$ @robjohn I think the fact, for which there is possibility to take a bounties points without solving of the problem it's a bad fact. It's my opinion only, of course and I think we need to delete this possibility from our forum. $\endgroup$ – Michael Rozenberg Jan 18 '17 at 12:36

This does not answer the question completely, but proves the inequalities for large enough $n$.

Let $f(x)=e^{g(x)}$ where $g(x)=\frac{1}{x}\log\Gamma(x+1)$, and $h(x)=\left(\frac{x}{x+1}\right)^x$.

It is enough to show that $$\frac{x}{x+1}h(x)<f'(x)<h(x),$$

since $f(n+1)-f(n)=f'(c_n)$ for some $n< c_n <n+1$ by the Mean Value Theorem.

We use a version of Stirling's approximation of $\Gamma$ function:

$$\log\Gamma(x+1)=x\log x-x+\frac{1}{2}\log(2\pi x)+\sum_{n=1}^\infty \frac{B_{2n}}{2n(2n-1)x^{2n-1}}$$

This yields the following asymptotic relations for large enough $x$.

$$g(x)=\log x-1+\frac{\log(2\pi x)}{2x}+O(\frac{1}{x^2}),$$ $$g'(x)=\frac{1}{x}+\frac{1}{2x^2}-\frac{\log(2\pi x)}{2x^2}+O(\frac{1}{x^3}),$$ $$f'(x)=e^{g(x)}g'(x)=\frac{1}{e}(1+\frac{1}{2x}-\frac{\log^2(2\pi x)}{8x^2}+O(\frac{\log x}{x^2})),$$ $$h(x)=\frac{1}{e}(1+\frac{1}{2x}+O(\frac{1}{x^2})),$$ $$\frac{x}{x+1}h(x)=\frac{1}{e}(1-\frac{1}{2x}+O(\frac{1}{x^2}))$$

Thus, we have the claim for large enough $x$, and this proves the inequalities for large enough $n$.

Remark1) Treating error terms extra carefully might give an explicit $N$ such that the inequalities hold for $n>N$.

Remark2) Once we find such $N$, we can check one by one for $n=1,2,\cdots N$.


By partial summation: $$\sum_{j=1}^{n}\log j = n\log n - n + \sum_{j=1}^{n}\left(1-j\log(1+1/j)\right),$$ so it is possible to fix @i707107's argument (removing "for any $N$ big enough") by noticing that $$\left(1-j\log(1+1/j)\right) \in \left(\frac{1}{2j},\frac{1}{2j+2}\right).$$

  • $\begingroup$ Can you please elaborate your answer on how to remove "$N$ big enough"? $\endgroup$ – Sungjin Kim Jan 28 '14 at 7:00

Here is a partial answer :

If a sequence $u=(u_n)_{n\ge1}$ of real numbers converges, then the sequence $\left(\frac{1}{n}\sum_{k=1}^nu_k\right)_{n\ge1}$ converges to the same limit. This is the well known Cesaro's lemma.

It can be proved that the converse is false (consider the sequence $u=((-1)^n)_{n\ge1}$) but becomes true if we assume that $u$ is monotonic.

As a consequence, we get the following result :

If $t=(t_n)_{n\ge1}$ is a monotonic sequence of real numbers such that $\lim_{n\to\infty}\frac{u_n}{n}=L\in\mathbb{R}$ then $\lim_{n\to\infty}\left(u_{n+1}-u_n\right)=L$.

Let's apply this last result to the sequence $t=(\left[n!\right])^{1/n})_{n\ge1}$.

It's easy to show ($\color{red}{\mathrm{see}\,\,\mathrm{below}}$) that


Therefore, if we prove that $t$ is monotonic (at least ultimately, which is a sufficient condition), we will reach the conclusion that :


It is a consequence of Cesaro's lemma that if a sequence of positive real numbers $(x_n)_{n\ge1}$ verifies $\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=L$ some $L>0$ then $\lim_{n\to\infty}\left(x_n\right)^{1/n}=L$.

Applying this to sequence $x_n=\left(\frac{n!}{n^n}\right)$ we obtain $\lim_{n\to\infty}\frac{\left[n!\right]^{1/n}}{n}=\frac{1}{e}$

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    $\begingroup$ Does this add anything to the other partial answers that already use Stolz-Cesaro? $\endgroup$ – robjohn Jan 23 '17 at 18:37
  • $\begingroup$ IMHO it reaches the same conclusion by a slightly different way. The Stolz-Cesaro property should not be confused with the partial reciprocal of Cesaro lemma. $\endgroup$ – Adren Jan 23 '17 at 18:50

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