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Show that $\left(2n\right)!\leq \left(n\left(n+1\right)\right)^{n}$.

My attempt so far:

$\left(2n \right)!=1\cdot2\cdot...\cdot\left(2n-1\right)\cdot2n\leq\left(n\left(n+1\right)\right)^{n}=\left(n^{2}+n\right)^{n}$

This is where I am stuck. I know that his holds for some values of $n$, such as 2, but I am having a hard time proceeding from here. I am not sure if induction would be the best approach or if there would be a more elegant solution. Thanks in advance for any hints.

Chairman Meow
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6 Answers6

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For $1\leq i \leq n$ we have $i(2n-i+1) \leq n^{2}+n$. This is because $n^{2}-(2i-1)n+i^{2}-i =(n-(i-\frac1 2))^{2}+i^{2}-i- (i-\frac1 2)^{2}\geq 0$. [ I have used the fact that $|n-(i-\frac 12)| \geq \frac1 2$ which gives $(n-(i-\frac 12))^{2} \geq \frac1 4$].

Now just mulptly the inequalities $i(2n-i+1) \leq n^{2}+n$ for $i=1,2,...,n$.

Kavi Rama Murthy
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Ross Millikan has perhaps the slickest approach, but if one doesn’t see it, one can also prove the result by induction on $n$. We can check the cases $n=0,1,2$ by hand. Suppose that $n\ge 2$, and $(2n)!\le n^n(n+1)^n$.

$$\frac{(2(n+1))!}{(2n)!}=2(n+1)(2n+1)\,,$$

and

$$\frac{(n+1)^{n+1}(n+2)^{n+1}}{n^n(n+1)^n}=\frac{(n+1)(n+2)^{n+1}}{n^n}\,,$$

so we’re done if we can show that

$$2(n+1)(2n+1)\le\frac{(n+1)(n+2)^{n+1}}{n^n}\,,$$

i.e., that

$$\frac{4n+2}{n+2}\le\left(\frac{n+2}n\right)^n\,.$$

And this follows from the binomial theorem:

$$\begin{align*} \left(\frac{n+2}n\right)^n&=\left(1+\frac2n\right)^n=\sum_{k=0}^n\binom{n}k\left(\frac2n\right)^k\\ &\ge 1+n\cdot\frac2n+\frac{n(n-1)}2\cdot\frac4{n^2}\\ &=3+\frac{2n-2}n\\ &=5-\frac2n\\ &\ge 4\\ &\ge 4-\frac6{n+2}\\ &=\frac{4n+2}{n+2} \end{align*}$$

Brian M. Scott
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You are done, but should probably put some words in to support the $\le$ sign. Break the product on the left up into pairs of numbers, taking the highest and lowest off the run to make the pairs. There are $n$ pairs and each one is less than $n(n+1)$

Ross Millikan
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  • Thank you for the input. So would the pairs be something like this: $1\cdot 2n, 2 \cdot 2n-1,...,\left(2n-1\right)*2n$? – Chairman Meow Jan 24 '21 at 05:10
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    The first two are correct, but the last one is $n\cdot (n+1)$ as you are working your way in from each end. You should also have parentheses around $2n-1$ in the second one. – Ross Millikan Jan 24 '21 at 05:11
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    @ChairmanMeow you have to give a formal proof of the fact that the product of each pair does not exceed $n^{2}+n$. This is what I have done in my answer. – Kavi Rama Murthy Jan 24 '21 at 05:21
  • "You are done"? I think the OP merely wrote down the goal ... – Hagen von Eitzen Jan 24 '21 at 05:32
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$$\left(2n\right)!\leq \left(n\left(n+1\right)\right)^{n}$$ $$\log((2n)!) \leq n \log(n(n+1))$$ Using Stirling approximation for the lhs and Taylor for the rhs $$\text{lhs}=n (2 \log (n)-2+2\log (2))+\frac{1}{2} \log (4 \pi n)+\frac{1}{24 n}+O\left(\frac{1}{n^3}\right)$$ $$\text{rhs}=2 n \log (n)+1-\frac{1}{2 n}+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\text{rhs- lhs}= 2(1-\log (2))n+\left(1-\frac{1}{2} \log (4 \pi n)\right)-\frac{13}{24 n}+O\left(\frac{1}{n^2}\right)$$

Claude Leibovici
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The Simplest way is as fallows
You can write down the inequality as: $$ (2n)! \le (n^2+n)^n $$ Since $ n \ge 1 $: $$ 2n \le n^2+n $$ $$ (2n)^n \le (n^2+n)^n $$ Also: $$ (2n)! \le (2n)^n $$ So: $$ (2n)! \le (n^2+n)^n $$

Khosro Moeini
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  • I wanted to write exactly this answer. So +1. – Allawonder Jan 24 '21 at 06:07
  • However, I would suggest you remove the first sentence. It is not quite justifiable -- what's more, it doesn't serve any positive purpose. – Allawonder Jan 24 '21 at 06:09
  • -1 Can you explain why $ (2n)! \leq (2n) ^n$? For example, with $n =2 $, we have $ 24 < 16$ isn't a true statement. Does it hold for some large enough $n$? Stirling's approximation tells us that we'd likely want something involving $(2n) ^ { 2n}$, so I'd be surprised if $(2n)^n$ was sufficient. – Calvin Lin Jan 24 '21 at 19:47
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Since $n! < n(n/e)^n$, $(2n)! < 2n(2n/e)^{2n}$, so we want $2n(2n/e)^{2n} \lt (n^2+n)^n $ or $(2n)^{1/n}(2n/e)^2 \lt n^2+n $.

This is true if $(2n)^{1/n}(2n/e)^2 \lt n^2 $ or $(2n)^{1/n} \lt e^2/4 $.

Since $(2x)^{1/x}$ is decreasing for $x > e/2$ and $(2n)^{1/n} \lt e^2/4$ for $n = 3$, (since $(e^2/4)^3 =e^6/64 \gt 6.3$) we are done.

marty cohen
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