How to prove $\left(\frac{n}{n+1}\right)^{n+1}<\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}<\left(\frac{n}{n+1}\right)^n$

Question

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 $n\in \Bbb N^{+}.$ creat by( wang yong xi)

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.

Answer 1

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

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

Answer 2

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

Answer 3

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

Answer 4

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.

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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 $$ \left(\frac{n}{n+1}\right)^n-\left(\frac{n}{n+1}\right)^{n+1}\sim\frac1{en}\tag{15} $$

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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.