Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

While simplifying an inequality, this inequality was derived:

$${(n+2)!}< \left(n(\sqrt{2}-1)+\sqrt{2}\right)^{n+2},\quad\quad\quad\quad n\in \mathbb{N}$$

Do you have any idea to prove it? It is very close for small $n$.

share|cite|improve this question
Please don't use display math in titles; it takes up a lot of unnecessary space on the front page. – Rahul Feb 6 '13 at 17:11
I am are asking for any hint to solve it, however you labelled your question with the "induction" tag. – Matemáticos Chibchas Feb 6 '13 at 17:15
@MatemáticosChibchas: hint or proof with or without induction is welcome :) (However a direct proof is better) – user59671 Feb 6 '13 at 17:17
For zero it is false :) – uforoboa Feb 6 '13 at 17:24
$n\in \mathbb{N}$ – user59671 Feb 6 '13 at 17:25

4 Answers 4

up vote 1 down vote accepted

A completely different approach, that's why I don't simply edit my first (lousy) attempt:

We want to show that $$\tag1 n!<(\alpha n+\beta)^n\quad\text{for }n\ge 3,$$ where $\alpha=\sqrt 2-1\approx0.414$ and $\beta=\sqrt 2-2\alpha=2-\sqrt 2\approx0.586$. Assume we findan inequality $$ \tag2 n!\le (\gamma n)^n\quad\text{for }n\ge n_0.$$ Then we have shown $(1)$ for all $n$ with $n_0\le n\le \frac\beta{\gamma-\alpha}$ if $\gamma>\alpha$. Of course, if $(2)$ holds even for some $\gamma <\alpha$, then we have shown $(1)$ for all $n\ge n_0$. We can try to find good such $\gamma$.

By the AGM inequality, we have $$\prod_{k=r}^s k\le \left(\frac{r+s}2\right)^{s-r+1}.$$ With $r=1$ this gives us immediatyly $(2)$ with $\gamma=\frac12$, hence $(1)$ for $n\le 6$ because $\frac\beta{\frac12-\alpha}\approx6.8$.

Let $a$ be a number with $0<a<1$ such that $an$ is an integer. Then $$\tag3n!=\prod_{k=1}^{an-1}k\cdot \prod_{k=an}^nk\le \left(\frac{an}{2}\right)^{an}\left(\frac{(1+a)n}{2}\right)^{n-an}=\left(\frac n2a^a(1+a)^{1-a}\right)^n.$$ Let $\gamma=\frac{a^a(1+a)^{1-a}}2$. Numerically, if $0.19<a<0.19+\frac17$ (which can definitely be achieved if $n\ge 7$), then $\gamma<0.42$, which gives us $(1)$ for $n\le 101$ (because $\frac{\beta}{0.42-\alpha}\approx101.2$). These results alone may already be helpful for other methods that establish $(1)$ for $n$ "big enough". However, we could not possibly proceed significantly further up.

We can refine $(3)$: If $an-1\ge 7$, we have already seen that we can estimate $(an-1)!\le (0.42an)^{an}$ and obtain $$\tag4 n!\le (0.42an)^{an}\left(\frac{(1+a)n}2\right)^{n-an}=(\gamma n)^n $$ with $$\gamma = (0.42a)^a\left(\frac{1+a}2\right)^{1-a}.$$ If $0.3<a<0.6$ we find numerically that $\gamma<0.41<\alpha$. This shows $(1)$ for all $n\ge 27$ (so that $an-1\ge 7$).

In summary, $(1)$ holds for all $n\ge 3$.

Remark: Note that a bit of work is hidden in what appears above a "numerical findings": You need to show that the expressions for $\gamma$ have a unique local minimum for $a\in(0,1)$, which then allows us to conclude $\gamma(a)\le\max\{\gamma(x_1),\gamma(x_2)\}$ for all $a\in[x_1,x_2]$.

Edit: It seems like I have made an off-by-one error somewhere, but don't see where exactly right now ...

share|cite|improve this answer

Useful bounds on $n!$ (rather than asymptotics) are given here; in this case, you can use the fact that $$ n! \le e n^{n+1/2}e^{-n} $$ for all $n\ge 1$. Taking the $n$-th root of both sides, $$ \sqrt[n]{n!} \le \left(\frac{n}{e}\right)\left(e \sqrt{n}\right)^{1/n}. $$ Now, $$ \left(e\sqrt{n}\right)^{1/n}=\exp\left({\frac{1}{n}\ln (e\sqrt{n})}\right)=\exp\left({\frac{2+\ln n}{2n}}\right) \rightarrow 1, $$ decreasing monotonically from above; in particular it's less than $9/8$ for $n\ge 22$ (by a direct calculation).

You want to show that $$ \sqrt[n]{n!}<(\sqrt{2}-1)(n-2)+\sqrt{2}=(\sqrt{2}-1)n+(2-\sqrt{2}) $$ for all $n\ge 3$. The above calculation shows the stronger inequality $$ \sqrt[n]{n!}<\frac{9}{8e}n<(\sqrt{2}-1)n$$ for $n \ge 22$. The remaining cases, $n=3,4,\ldots,21,$ can then be shown by hand to satisfy the original inequality. As you point out, for $n=3$ and $n=4$ it is close (and of course for $n=1$ and $n=2$ you have equality).

share|cite|improve this answer
thanks. still a lot of calculations for n=3,...,21 is needed! but mathematically logical. – user59671 Feb 6 '13 at 19:10

Using Stirling, $$n!\approx n^ne^{-n}\sqrt {2\pi n}$$ we can approximate the left hand side with $$ (n+2)!\approx (n+2)^{n+2}e^{-n-2}\sqrt{2\pi (n+2)}$$ and the right hand side with $$ (n(\sqrt 2-1)+\sqrt2)^{n+2}>n^{n+2}(\sqrt2-1)^{n+2}=(n+2)^{n+2}\left(1+\frac2n\right)^{-(n+2)}(\sqrt2-1)^{n+2}.$$ If we drop the common factor $(n+2)^{n+2}$ and note that $\left(1+\frac2n\right)^{-(n+2)}=\left(1-\frac2{n+1}\right)^{n+2}\to e^{-2}$, we need only compare the growth of $\sqrt{2\pi (n+2)}$ and $((\sqrt 2-1)e)^{n+2}$ . Since $(\sqrt 2-1)e>1$, the exponential "wins" for $n$ big enough. To turn this into a proper proof, you need to take a precise look at the error hidden in the "$\approx$" of Stirling and tthus determine, what "$n$ big enough" means. It turns out that the relative error is below $\frac 1{12n}$ and therefore $n=22$ can be taken as "big enough".

EDIT: Thanks to CutieKrait's comment, I noted that I had $(1+\frac2n)^{n+2}$ instead of its reciprocal. Thus one needs to have a closer look at its value as well (it's simply no longer "$>1$", but at least bounded from below ...)

share|cite|improve this answer
why $$n^{n+2}(\sqrt2-1)^{n+2}=(n+2)^{n+2}\left(1+\frac2n\right)^{n+2}(\sqrt2-1)^{n+2‌​}$$? And you used a few approximations. finding a large enough $n$ and proving for other $n$s may be harder. – user59671 Feb 6 '13 at 17:58
It's funny that my approach also hit on $n=22$ as a cutoff... a coincidence, I think, but I'm not sure. – mjqxxxx Feb 6 '13 at 18:53

Let $$a_k=\sqrt[k] {k!}$$ By Prove elementarily that $\sqrt[n+1] {(n+1)!} - \sqrt[n] {n!}$ is strictly decreasing we have $$a_{k+2}-a_{k+1}< a_{k+1}-a_k$$ taking sum we get this inequality: $$\sum_{k=1}^{m}{a_{k+2}-a_{k+1}}< \sum_{k=1}^{m}{a_{k+1}-a_k}$$ that is $$a_{m+2}-a_{2}< a_{m+1}-a_1$$ that is $$a_{m+2}-a_{m+1}< a_{2}-a_1=\sqrt{2}-1$$ taking sum again we get this inequality: $$\sum_{m=1}^{n}{a_{m+2}-a_{m+1}}< n(\sqrt{2}-1)$$ that is:

$$a_{n+2}-a_{2}<n(\sqrt{2}-1)$$ or $$a_{n+2}<n(\sqrt{2}-1)+\sqrt{2}$$ or $$\sqrt[n+2] {(n+2)!}< n(\sqrt{2}-1)+\sqrt{2}$$ or $${(n+2)!}< \left(n(\sqrt{2}-1)+\sqrt{2}\right)^{n+2}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.