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Arrange the following numbers in decreasing order: $$(300)^{600} \qquad 600! \qquad (200)^{600}$$

My progress so far

$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}+\cdots $$

$$e^x>\frac{x^n}{n!}$$ for $x=n$

$$n!>\bigg(\frac{n}{e}\bigg)^n>\bigg(\frac{n}{3}\bigg)^n$$

For $n=600$, I am having $600!>(200)^{600}$.

Please show me how I can prove the other inequalities.

Blue
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rekha pimoli
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    This is not a rigorous answer, but$$600!=1\cdot2\cdot3\cdot4\cdot\cdot\cdot600=\underbrace{(1\cdot600)\cdot(2\cdot599)\cdot(3.598)\cdot\cdot\cdot(300\cdot301)}_{300\text{ terms}}$$Each term is of the form $n(601-n)$ which peaks for $n=300$ and decreases when we go away.

    Only the last couple terms, i.e. $(300\cdot301),(299\cdot302),...,(284\cdot317)$, are greater than $300^2$ and the vast majority of terms in the product are less than $300^2$.$$\implies\underbrace{(1\cdot600)\cdot(2\cdot599)\cdot(3.598)\cdot\cdot\cdot(300\cdot301)}_{300\text{ terms}}\le(300^2)^{300}$$

     – Shubham Johri Apr 06 '19 at 12:08
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    May be, you could use logarithms for the first and third and Stirling approximation of $\log(n!)$ for the second one. – Claude Leibovici Apr 06 '19 at 14:23
  • I think just remember that $\sqrt[n]{n!} \approx n/e$, then take the $600$-th root of each, and conclude that $200^{600} < 600! < 300^{600}$. – mjqxxxx Oct 07 '24 at 22:54

3 Answers3

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$$\frac{600!}{300^{600}}=2\prod^{299}_{r=1}\frac{(300-r)(300+r)}{(300)^2}$$

$$\frac{600!}{(300)^{600}}=2\prod^{299}_{r=1}\bigg(1-\frac{r^2}{(300)^2}\bigg)<1$$

So we have $$(600)!<(300)^{600}$$

DXT
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We have, for $n\ge 6$, $$\bigg(\frac n2\bigg)^n\gt n!\gt \bigg(\frac n3\bigg)^n\tag1$$ So, for $n=600$, we obtain $$300^{600}\gt 600!\gt 200^{600}$$

To prove $(1)$, induction works.

For $n=6$, $(1)$ holds.

In the following, we use the fact that for $n\ge 2$, $$2\lt \bigg(1+\frac 1n\bigg)^n\lt 3$$

(We have $(1+\frac 1n)^n\gt \binom n0(\frac 1n)^0+\binom n1(\frac 1n)^1=2$, and see for example this proof for the inequality $(1+\frac 1n)^n\lt 3$.)

Suppose that $(1)$ holds for $n$.

Then, we have $$\begin{align}&\bigg(\frac{n+1}{2}\bigg)^{n+1}-(n+1)! \\\\&=\bigg(\frac{n+1}{2}\bigg)^{n+1}-(n+1)\times n! \\\\&\gt \bigg(\frac{n+1}{2}\bigg)^{n+1}-(n+1)\times \bigg(\frac n2\bigg)^n \\\\&=\frac{n+1}{2}\bigg(\frac n2\bigg)^n\bigg(\underbrace{\bigg(1+\frac 1n\bigg)^n-2}_{\text{positive}}\bigg) \\\\&\gt 0\end{align}$$

and

$$\begin{align}&(n+1)!-\bigg(\frac{n+1}{3}\bigg)^{n+1} \\\\&=(n+1)\times n!-\bigg(\frac{n+1}{3}\bigg)^{n+1} \\\\&\gt (n+1)\times \bigg(\frac n3\bigg)^n-\bigg(\frac{n+1}{3}\bigg)^{n+1} \\\\&=\frac{n+1}{3}\bigg(\frac n3\bigg)^n\bigg(\underbrace{3-\bigg(1+\frac 1n\bigg)^n}_{\text{positive}}\bigg) \\\\&\gt 0.\ \blacksquare\end{align}$$

mathlove
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$$\ln(n!)=\sum_{k=1}^n\ln k<\int_1^{n+1}\ln xdx=(n+1)\ln(n+1)-n$$

$$A=\ln(600!)<601\ln601-600$$ $$B=\ln(300^{600})=600\ln300$$ $$B>A$$ if $$600\ln300>601\ln601-600$$ if $$1>\ln\left(\frac{601}{300}\right)+\frac{\ln601}{600}$$ if $$1>\ln2+\ln(1+\frac1{600})+\frac{10}{600}$$ if $$1>0.694+\frac1{600}+\frac1{60}$$ if $$1>0.694+0.0017+0.017$$ which is true.

Bob Dobbs
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