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I tried to prove that if $n>2$ then $(n!)^2>n^n$ but did not managed. That is the trick to compare those as both grows rapidly? Induction seems hard: $((n+1)!)^2=(n+1)^2(n!)^2>(n+1)^2n^n$ but why $(n+1)^2n^n>(n+1)^{n+1}$? I also noted that $n^n=e^{n\ln n}$ and $n!^2=e^{2\ln n!}=e^{2\sum_{i=1}^n \ln i}$ but got stuck.

novice
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2 Answers2

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Note:

$$n!^2 = (1\cdot n)\cdot (2\cdot(n-1))\cdots(n\cdot 1)$$

Prove that $k\cdot (n+1-k)\geq n$, for $k=1,2,\dots, n$. So $$n!^2\geq n^n$$

With $n>2$, we have $2\cdot (n-1)>n$, so it is strict inequality when $n>2$.

Thomas Andrews
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  • Do you use induction & the symmetry of the function to prove the lemma or is there a better way? – Colm Bhandal Nov 30 '15 at 21:02
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    I'd use calculus - it follows from the shape of $f(x)=x(1-x)$, which is increasing when $x<1/2$ and decreasing when $x>1/2$. Use that to show that $f\left(\frac{1}{n+1}\right)\leq f\left(\frac{k}{n+1}\right)$, then multiply by $(n+1)^2$. But I'm sure there are more direct ways to prove it. – Thomas Andrews Nov 30 '15 at 21:07
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    For $k$ between $1$ and $n$, $((n+1)-k)\cdot k = \left(\frac{n+1}{2 }+\frac{n-2k+1}{2}\right)\left(\frac{n+1}{2}-\frac{n-2k+1}{2 }\right)=\left(\frac{n+1}{2}\right)^2-\left(\frac{n-2k+1}{2} \right)^2\ge\left(\frac{n+1}{2}\right)^2-\left(\frac{n-1}{2}\right)^2=n$, and the inequality is strict if $k$ is strictly between $1$ and $n$. – Steve Kass Nov 30 '15 at 21:59
  • Nice. Yeah my method was slightly different using induction on $k$. $(k + 1)(n + 1 - (k+1)) = k(n + 1 - k) + n - 2k$. For all $k \leq \frac{n}{2}$ this works by induction. But since the function is symmetric around $\frac{n}{2}$, we immediately get that the inequality also holds for all $k \geq \frac{n}{2}$ also. – Colm Bhandal Dec 01 '15 at 13:38
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Suppose $(n!)^2 > n^n$. Then $((n+1)!)^2 =(n!)^2(n+1)^2 >n^n(n+1)^2 $ so we need $n^n(n+1)^2 \ge (n+1)^{n+1} $ which is the same as $n^n \ge (n+1)^{n-1} $ or, dividing by $n^{n-1}$, $n \ge (1+1/n)^{n-1} $. Multiplying by $1+1/n$, this becomes $n+1 \ge (1+1/n)^n $.

This is true because $(1+1/n)^n < e $, as has been shown many times here. Here is one of the easier:

Using the binomial theorem,

$\begin{array}\\ (1+1/n)^n &=\sum_{k=0}^n \binom{n}{k}(1/n)^k\\ &=\sum_{k=0}^n \frac{n!}{k!(n-k)!}(1/n)^k\\ &=\sum_{k=0}^n \frac{\prod_{j=0}^{k-1}(n-j)}{k!n^k}\\ &=\sum_{k=0}^n \frac{\prod_{j=0}^{k-1}(1-j/k)}{k!}\\ &<\sum_{k=0}^n \frac{1}{k!}\\ &<\sum_{k=0}^{\infty} \frac{1}{k!}\\ &= e \end{array} $

Actually, all we need is a bound on $\sum_{k=0}^{\infty} \frac{1}{k!} $. An easy one is gotten from $k! \ge 2\cdot 2^{k-2}$ for $k \ge 2$, easily proved by induction. Then

$\begin{array}\\ \sum_{k=0}^{\infty} \frac{1}{k!} &=1+1+\sum_{k=2}^{\infty} \frac{1}{k!}\\ &\lt 2+\sum_{k=2}^{\infty} \frac{1}{2\cdot 2^{k-2}}\\ &\lt 2+\frac1{2}\sum_{k=0}^{\infty} \frac{1}{2^k}\\ &= 2+1\\ &= 3 \end{array} $

marty cohen
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