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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.

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  • $\begingroup$ 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
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Evaluating the Limit

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.


Asymptotic Expansions

This is not what I would call simple, but it does show that, at least asymptotically, the initial inequality is true.

Using the Euler-Maclaurin Sum Formula, we get the formula $$ \log(n!)=\frac12\log(2\pi n)+n\log(n)-n+\frac1{12n}-\frac1{360n^3}+\frac1{1260n^5}+O\!\left(\frac1{n^7}\right)\tag7 $$ and therefore, $$ \frac1n\log(n!)=\log(n)-1+\frac12\frac{\log(2\pi n)}n+\frac1{12n^2}-\frac1{360n^4}+\frac1{1260n^6}+O\!\left(\frac1{n^8}\right)\tag8 $$ The constant $\frac12\log(2\pi)$ is gotten elsewhere (e.g. see this answer).

Using $\log(n+1)=\log(n)+\frac1n-\frac1{2n^2}+\frac1{3n^3}-\frac1{4n^4}+\dots$, we get $$ \log((n+1)!)=\frac12\log(2\pi n)+(n+1)\log(n)-n+\frac{13}{12n}-\frac1{2n^2}+O\!\left(\frac1{n^3}\right)\tag9 $$ and therefore, using $\frac1{n+1}=\frac1n-\frac1{n^2}+\frac1{n^3}-\frac1{n^4}+O\!\left(\frac1{n^5}\right)$, we get $$ \scriptsize\frac1{n+1}\log((n+1)!)=\log(n)-1+\frac12\frac{\log(2\pi n)}{n}+\frac1n-\frac12\frac{\log(2\pi n)}{n^2}+\frac1{12n^2}+O\!\left(\frac{\log(n)}{n^3}\right)\tag{10} $$ Exponentiating $(8)$ and $(10)$ and subtracting gives $$ \bbox[5px,border:2px solid #C0A000]{\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}=\frac1e\left(1+\frac1{2n}-\frac{5+3\log(2\pi n/e)^2}{24n^2}+O\!\left(\frac{\log(n)^3}{n^3}\right)\right)}\tag{11} $$ Using $\log\left(\frac{n}{n+1}\right)=-\frac1n+\frac1{2n^2}-\frac1{3n^3}+O\!\left(\frac1{n^4}\right)$ yields $$ \bbox[5px,border:2px solid #C0A000]{\left(\frac{n}{n+1}\right)^n=\frac1e\left(1+\frac1{2n}-\frac5{24n^2}+O\!\left(\frac1{n^3}\right)\right)}\tag{12} $$ and $$ \bbox[5px,border:2px solid #C0A000]{\left(\frac{n}{n+1}\right)^{n+1}=\frac1e\left(1-\frac1{2n}+\frac7{24n^2}+O\!\left(\frac1{n^3}\right)\right)}\tag{13} $$ Thus, asymptotically, $$ \left(\frac{n}{n+1}\right)^n-\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)\sim\frac{\log(2\pi n/e)^2}{8n^2}\tag{14} $$ which is far smaller than $$ \left(\frac{n}{n+1}\right)^n-\left(\frac{n}{n+1}\right)^{n+1}\sim\frac1{en}\tag{15} $$


Graphical Comparisons

Graphically, we can see the size difference between $(14)$, the distance between the top two functions, and $(15)$, the distance between the top and bottom functions.

enter image description here

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  • $\begingroup$ 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
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    $\begingroup$ @MichaelRozenberg: the bounty I received two years ago, was given because someone liked the progress made, not necessarily for a complete answer. The bounty offered now is spurring people to work on this problem again. $\endgroup$ – robjohn Jan 19 '17 at 0:37
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Hint:

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
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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 $n=1,2,\cdots N$.

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  • $\begingroup$ Could you go from large $x$ back to smaller $x$? In principle yes, but I do not know how difficult it would be to show that a certain derivative is negative (if it is...might be naive to expect this would be any easier than the original problem, but I am curious). For example, let $L(x)=\sqrt[x+1]{\Gamma(x+2)}-\sqrt[x]{\Gamma(x+1)}-(\frac x{x+1})^{x+1}$ , for $x>0$. If $L'(x)<0$ for all $x>0$ and if $L(x)>0$ for large enough $x$ then $L(x)>0$ for all $x>0$. Wolfram alpha gave a plot of $L'(x)$ which looks like a negative function. I posted an answer contemplating this approach $\endgroup$ – Mirko Sep 30 '19 at 16:04
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Remark: I proved the second inequality. Maybe the first inequality can be proved in similar way. I have not yet tried it. I used Maple software.

Let us prove that $$\sqrt[n+1]{(n+1)!} - \sqrt[n]{n!} < \Big(\frac{n}{n+1}\Big)^n, \quad \forall n\ge 1.$$

We first introduce some auxiliary results (Facts 1 through 6) used to prove the inequality.

Fact 1: For all integers $n\ge 1$, we have \begin{align} \sqrt[n]{n!} &> \sqrt[2n]{\pi}\, \frac{n}{\mathrm{e}} \sqrt[6n]{8n^3 + 4n^2 + n + \frac{1}{100}}, \\ \sqrt[n]{n!} &< \sqrt[2n]{\pi}\, \frac{n}{\mathrm{e}} \sqrt[6n]{8n^3 + 4n^2 + n + \frac{1}{30}}, \\ \sqrt[n+1]{(n+1)!} &< \sqrt[2n+2]{\pi}\, \frac{n+1}{\mathrm{e}} \sqrt[6n+6]{8n^3 + 28n^2 + 33n + \frac{391}{30}}. \end{align} See: G. E. Andrews and B. C. Berndt, Ramanujan's Lost Notebook Part IV, 2013, page 111.

Fact 2: For all integers $n\ge 1$, we have $$\ln \Big(8n^3 + 28n^2 + 33n + \frac{391}{30}\Big) - \ln \Big(8n^3 + 4n^2 + n + \frac{1}{100}\Big) < \frac{3}{n} - \frac{2}{n^2} + \frac{3607}{2400n^3}.$$ The proof is simple and thus omitted.

Fact 3: For all integers $n\ge 1$, we have $$\ln (8n^3 + 4n^2 + n + \frac{1}{100}) > 3\ln 2 + 3\ln n + \frac{1}{2n} - \frac{47}{2400n^3}.$$ The proof is simple and thus omitted.

Fact 4: For all integers $n\ge 1$, we have $$\ln (8n^3 + 4n^2 + n + \frac{1}{30}) < 3\ln 2 + 3\ln n + \frac{1}{2n}.$$ The proof is simple and thus omitted.

Fact 5: For all integers $n\ge 1$, we have $$n\ln n - n \ln (n+1) > -1 + \frac{1}{2n} - \frac{1}{3n^2}.$$ The proof is simple and thus omitted.

Fact 6: For all integers $n\ge 1$, we have $$\sqrt[n+1]{(n+1)!} < \Big(1 + \frac{1}{n} - \frac{\ln n + \ln 2\pi - 1}{2n^2}\Big)\sqrt[n]{n!}.$$ The proof is given later.

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Now let us begin. If $1 \le n\le 5$, the inequality is verified directly. In the following, we assume that $n \ge 6$.

According to Fact 6, it suffices to prove that $$\Big(\frac{1}{n} - \frac{\ln n + \ln 2\pi - 1}{2n^2}\Big)\sqrt[n]{n!} < (\frac{n}{n+1})^n.$$ According to Fact 1, it suffices to prove that \begin{align} &\ln \Big(\frac{1}{n} - \frac{\ln n + \ln 2\pi - 1}{2n^2}\Big) + \frac{1}{2n}\ln \pi + \ln n - 1 + \frac{1}{6n}\ln (8n^3 + 4n^2 + n + \frac{1}{30})\\ <\ & n\ln n - n \ln (n+1). \end{align} According to Fact 4 and 5, it suffices to prove that \begin{align} &\ln \Big(\frac{1}{n} - \frac{\ln n + \ln 2\pi - 1}{2n^2}\Big) + \frac{1}{2n}\ln \pi + \ln n - 1 + \frac{1}{6n}(3\ln 2 + 3\ln n + \frac{1}{2n})\\ <\ & -1 + \frac{1}{2n} - \frac{1}{3n^2}. \end{align} Let \begin{align} G(x) &= -1 + \frac{1}{2x} - \frac{1}{3x^2} - \ln \Big(\frac{1}{x} - \frac{\ln x + \ln 2\pi - 1}{2x^2}\Big)\\ &\quad - \Big(\frac{1}{2x}\ln \pi + \ln x - 1 + \frac{1}{6x}(3\ln 2 + 3\ln x + \frac{1}{2x})\Big). \end{align} We have $$G'(x) = \frac{1}{6x^3(2x - \ln x - \ln 2\pi + 1)}(p_2(\ln x)^2 + p_1 \ln x + p_0)$$ where \begin{align} p_2 &= -3x, \\ p_1 &= -(6\ln 2\pi - 9)x - 5, \\ p_0 &= (-3\ln^2 2\pi + 9\ln 2\pi +4)x + 5 - 5\ln 2\pi. \end{align} Clearly, $p_2, p_1 < 0$ for $x \ge 1$. Since $\ln x > \frac{2(x-1)}{x+1}$ for $x > 1$, we have $$p_2(\ln x)^2 + p_1 \ln x + p_0 < p_2\Big(\frac{2(x-1)}{x+1}\Big)^2 + p_1\frac{2(x-1)}{x+1} + p_0, \quad \forall x \ge 6.$$ It is easy to prove that $$p_2\Big(\frac{2(x-1)}{x+1}\Big)^2 + p_1\frac{2(x-1)}{x+1} + p_0 < 0, \quad \forall x \ge 6.$$ Thus, $G'(x) < 0$ for $x\ge 6$. Note also that $\lim_{x\to \infty} G(x) = 0$. Thus, we have $G(x) > 0$ for $x\ge 6$. This completes the proof.

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Proof of Fact 6: If $1 \le n\le 3$, the inequality is verified directly. In the following, we assume that $n \ge 4$. We need to prove that \begin{align} \ln \sqrt[n+1]{(n+1)!} < \ln \sqrt[n]{n!} + \ln \Big(1 + \frac{1}{n} - \frac{\ln n + \ln 2\pi - 1}{2n^2}\Big). \end{align} According to Fact 1, it suffices to prove that \begin{align} &\frac{1}{2n+2}\ln \pi + \ln (n+1) - 1 + \frac{1}{6n+6}\ln (8n^3 + 28n^2 + 33n + \frac{391}{30})\\ < \ & \frac{1}{2n}\ln \pi + \ln n - 1 + \frac{1}{6n}\ln (8n^3 + 4n^2 + n + \frac{1}{100})\\ &\quad + \ln \Big(1 + \frac{1}{n} - \frac{\ln n + \ln 2\pi - 1}{2n^2}\Big) \end{align} or \begin{align} & \frac{1}{6n+6}\Big(\ln (8n^3 + 28n^2 + 33n + \frac{391}{30}) - \ln (8n^3 + 4n^2 + n + \frac{1}{100})\Big)\\ < \ & \frac{1}{2n(n+1)}\ln \pi + \frac{1}{6n(n+1)}\ln (8n^3 + 4n^2 + n + \frac{1}{100})\\ &\quad + \ln \Big(1 - \frac{\ln n + \ln 2\pi - 1}{2n(n+1)}\Big). \end{align} According to Fact 2 and 3, it suffices to prove that \begin{align} & \frac{1}{6n+6}\Big(\frac{3}{n} - \frac{2}{n^2} + \frac{3607}{2400n^3}\Big)\\ < \ & \frac{1}{2n(n+1)}\ln \pi + \frac{1}{6n(n+1)} \Big(3\ln 2 + 3\ln n + \frac{1}{2n} - \frac{47}{2400n^3}\Big)\\ &\quad + \ln \Big(1 - \frac{\ln n + \ln 2\pi - 1}{2n(n+1)}\Big). \end{align} Let \begin{align} F(x) &= \frac{1}{2x(x+1)}\ln \pi + \frac{1}{6x(x+1)}\Big(3\ln 2 + 3\ln x + \frac{1}{2x} - \frac{47}{2400x^3}\Big)\\ &\quad + \ln \Big(1 - \frac{\ln x + \ln 2\pi - 1}{2x(x+1)}\Big) - \frac{1}{6x+6}\Big(\frac{3}{x} - \frac{2}{x^2} + \frac{3607}{2400x^3}\Big). \end{align} We have $$F'(x) = \frac{1}{3600x^5(x+1)^2(2x^2-\ln x - \ln 2\pi + 2x+1)}(q_2(\ln x)^2 + q_1\ln x + q_0)$$ where \begin{align} q_2 &= 3600x^4+1800x^3, \\ q_1 &= (7200\ln 2\pi - 9000)x^4 + (3600\ln 2\pi - 900)x^3 - 607x^2 - 2764x - 47, \\ q_0 &= -9000x^5 + (3600\ln^2 2\pi - 9000\ln 2\pi -2386)x^4 +(1800\ln^2 2\pi - 900\ln 2\pi+5842)x^3\\ &\quad +(-607\ln 2\pi + 6229)x^2 + (-2764\ln 2\pi + 2858)x - 47\ln 2\pi +47. \end{align} It is easy to prove that $q_2, q_1 > 0$ for $x\ge 1$. Since $x > \ln^2 x$ and $x > \ln x$ for $x\ge 1$, we have $$q_2(\ln x)^2 + q_1\ln x + q_0 < q_2 x + q_1 x + q_0, \quad \forall x\ge 4.$$ It is easy to prove that $q_2 x + q_1 x + q_0 < 0$ for $x \ge 4$. Thus, we have $F'(x) < 0$ for $x\ge 4$. Note also that $\lim_{x\to \infty} F(x) = 0$. Thus, we have $F(x) > 0$ for $x \ge 4$. This completes the proof of Fact 6.

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  • $\begingroup$ wow, i tried to prove the second inequality and quickly realized it would involve very, very precise estimates. im surprised someone did them. how long did this take you? $\endgroup$ – mathworker21 Oct 2 '19 at 0:01
  • $\begingroup$ Yes, the difficulty arises from two aspects: First, we need good bounds for all integers. Second, when we want to prove that $f(x) > 0$ where $f(x)$ is the form (seems complicated) of e.g., G(x) and F(x) in my proof, actually it can be easily proved by calculus. I took some hours. $\endgroup$ – River Li Oct 2 '19 at 0:37
  • $\begingroup$ Very interesting. I definitely learned something from this. I was looking for upper and lower bounds of $\sqrt[n]{n!}$ incorporating $n/e$ that were asymptotically similar and that was just the first part of your answer. $\endgroup$ – user123641 Oct 2 '19 at 13:34
  • $\begingroup$ @Robert Wolfe I also tried weaker bounds but they are not enough. The bounds can be found in en.wikipedia.org/wiki/Stirling%27s_approximation $\endgroup$ – River Li Oct 2 '19 at 15:06
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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

$$\lim_{n\to\infty}\frac{\left[n!\right]^{1/n}}{n}=\frac{1}{e}$$

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

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


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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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 $$\left(1-j\log(1+1/j)\right) \in \left(\frac{1}{2j},\frac{1}{2j+2}\right).$$

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  • $\begingroup$ Can you please elaborate your answer on how to remove "$N$ big enough"? $\endgroup$ – Sungjin Kim Jan 28 '14 at 7:00
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Could one go from large $x$ (the case already treated in @SungjinKim 's answer) back to smaller $x$? In principle yes, but I do not know how difficult it would be to show that a certain derivative is negative (if it is...might be naive to expect this would be any easier than the original problem, but I am curious). For example, let
$L(n)=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}-(\frac n{n+1})^{n+1}$ and
$L(x)=\sqrt[x+1]{\Gamma(x+2)}-\sqrt[x]{\Gamma(x+1)}-(\frac x{x+1})^{x+1}$ , for $x>0$.
If $L'(x)<0$ for all $x>0$ and if $L(x)>0$ for large enough $x$ then $L(x)>0$ for all $x>0$ (which would show the first of the two OP inequalities $(\frac n{n+1})^{n+1}<\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\ $).

I didn't do much re this new question (not an answer), apart from asking wolfram alpha, which gave a plot of $L'(x)$ which looks like a negative function indeed.

https://www.wolframalpha.com/input/?i=derivative+of+%28+%28Gamma%28x%2B2%29%29%5E%281%2F%28x%2B1%29%29-%28Gamma%28x%2B1%29%29%5E%281%2Fx%29-%28x%2F%28x%2B1%29%29%5E%28x%2B1%29++++%29+for+x+from+0+to+60

Just for the sake of avoiding $x+2$ in the expression (at the expence of introducing $x-1$), one may also wish to consider
$G(x)=L(x-1)=\sqrt[x]{\Gamma(x+1)}-\sqrt[x-1]{\Gamma(x)}-(\frac{x-1}x)^x$ , for $x>1$

Similarly, one may introduce
$R(x)=(\frac x{x+1})^x-\sqrt[x+1]{\Gamma(x+2)}+\sqrt[x]{\Gamma(x+1)}$
and try to show that its derivative is negative for $x>0$.
(Or work instead with $R(x-1)=(\frac{x-1}x)^{x-1}-\sqrt[x]{\Gamma(x+1)}+\sqrt[x-1]{\Gamma(x)}$, for $x>1$).

"Confirmed" by wolfram alpha for $0<x<60$:
https://www.wolframalpha.com/input/?i=derivative+of+%28+%28x%2F%28x%2B1%29%29%5Ex-%28Gamma%28x%2B2%29%29%5E%281%2F%28x%2B1%29%29%2B%28Gamma%28x%2B1%29%29%5E%281%2Fx%29+++++%29+for+x+from+0+to+60

Computing some values (with Reduce computer algebra) it seems that the second of the two inequalities (i.e. $\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}<(\frac n{n+1})^n\ $) is a bit tighter than the first one ($(\frac n{n+1})^{n+1}<\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\ $), (but that both hold true).

I feel I have no idea what I am talking about (and that I didn't do much by just posting a couple of links to wolfram alpha) so I post this as community wiki. (I read a little bit about the derivative of $\Gamma(x)$ on wikipedia and it looks like I need to learn a bit more to be able to follow.)

Also from wolfram alpha:
https://www.wolframalpha.com/input/?i=derivative+of+%28+%28x%2F%28x%2B1%29%29%5Ex-%28Gamma%28x%2B2%29%29%5E%281%2F%28x%2B1%29%29%2B%28Gamma%28x%2B1%29%29%5E%281%2Fx%29+++++%29+

$\frac d{dx}\bigl((\frac x{x + 1})^x-\Gamma(x+2)^{\frac 1{x + 1}}+\Gamma(x+1)^{\frac1x}\bigr)=$
${(\Gamma(x+1))}^{\frac1x} \bigl( \frac{\psi^{(0)}(x+1)}x- \frac{\log(\Gamma(x+1))}{x^2} \bigr)+$
$\ (\frac x{x + 1})^x \bigl( (x+1) (\frac1{x + 1}-\frac x{(x+1)^2})+ \log(\frac x{x + 1}) \bigr)-$
$\ \ (\Gamma(x+2))^{\frac 1{x + 1}}\bigl(\frac{\psi^{(0)}(x+2)}{x+1}- \frac{\log(\Gamma(x + 2))}{(x + 1)^2}\bigr)$

$\endgroup$

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