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Proving $$(1)\;\;\;\;\left(\frac n3\right)^n<n!<\left(\frac n2\right)^n\forall\,\,n\ge 6,n\in\mathbb{N}$$

and $$(2)\;\; \sqrt{n^n}\le n!\,\,\forall n\in\mathbb{N}$$ without induction

$\bf{My\; Try::}$ For $\bf{R.H.S}$ Inequality.

Using $\bf{A.M\geq G.M\;,}$ We get $$\frac{2+3+4+5+......+n}{n-1}>(2\cdot 3\cdot 4\cdot 5\cdot.......n)^{n-1}$$

So we get $$\frac{(n-1)(n+2)}{2(n-1)}>(n!)^{n-1}\Rightarrow n!<\left(\frac{n+2}{2}\right)^{n-1}$$

Now How can I solve after that also How can I prove Second one

Thanks

Community
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juantheron
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1 Answers1

4

For (2), note that if $n$ is even we have:

$$\begin{align*}\dfrac{n^n}{(n!)^2}&=\dfrac{n\cdot n\cdot n\cdot n\cdots n\cdot n}{n\cdot n\cdot (n-1)\cdot (n-1)\cdots2\cdot 2\cdot 1\cdot 1}\\&=\dfrac{n\cdot n}{n\cdot n}\cdot \dfrac{n\cdot n}{(n-1)(n-1)}\cdots \dfrac{n\cdot n}{(n/2+1)(n/2+1)}\cdot \dfrac{1}{(n/2)(n/2)}\cdots \dfrac{1}{1\cdot 1}\\ &=\prod_{k=1}^{n/2}\dfrac{n^2}{(n-k+1)^2}\cdot \prod_{k=1}^{n/2} \dfrac{1}{k^2}=\prod_{k=1}^{n/2}\left(\dfrac{n}{n-k+1}\right)^2\cdot \prod_{k=1}^{n/2} \dfrac{1}{k^2}\\ &=\prod_{k=1}^{n/2}\left(1+\dfrac{k-1}{n-k+1}\right)^2\cdot \prod_{k=1}^{n/2} \dfrac{1}{k^2}=\prod_{k=1}^{n/2} \left[ \dfrac{1}{k^2}\left(1+\dfrac{k-1}{n-k+1}\right)^2\right]\\ &=\prod_{k=2}^{n/2} \left[ \dfrac{1}{k^2}\left(1+\dfrac{k-1}{n-k+1}\right)^2\right] \hspace{2cm}\text{(for $k=1$, the factor is $1$)}\\ &=\prod_{k=2}^{n/2} \left[\dfrac{1}{k}\left(1+\dfrac{k-1}{n-k+1}\right)\right]^2 \end{align*}$$ In the case when $n$ is odd, the limits in the product change from $1$ to $\lfloor n/2\rfloor$ and an extra factor of $\dfrac{n}{\left(n-\lfloor n/2 \rfloor \right)^2}\leq \dfrac{n}{(n/2)^2}\leq \dfrac{4}{n}$ must be multiplied at the end.

In any case, since $\dfrac{1}{k}\leq \dfrac{1}{2}$ and $1+\dfrac{k-1}{n-k+1}\leq 2$ for each $k$, we have that each factor is less than or equal to $1$, and so $$\dfrac{n^n}{(n!)^2}\leq 1 \Rightarrow \sqrt{n^n}\leq n!$$

Darío G
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