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Which one is greater $600!$ or $300^{600}$

$\bf{My\; Try::}$ I have used Stirling Approximation.

For large $n>2\;,$ We can write $\displaystyle n! \approx \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$

So $$600!\approx \left(\frac{600}{e}\right)^{600}\sqrt{2\cdot \pi \cdot 600}<300^{600}$$

My question is how can we i solve using algebraic Inequalities , Help required

Thanks

juantheron
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3 Answers3

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Divide them , you will notice $$\frac{600!}{300^{600}}=2\prod_{k=1}^{299}\frac{(300-k)(300+k)}{300^2}=2\prod_{k=1}^{299}\left(1-\frac{k^2}{300^2}\right)$$ Now take $k =299$ out , you get $$\frac{600!}{300^{600}}=2\left(1-\frac{299^2}{300^2}\right)\prod_{k=1}^{298}\left(1-\frac{k^2}{300^2}\right)$$ Now we are multiplying a lot of terms but all of them are less than $1$ so $600!<300^{600}$.

Did
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avz2611
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  • Very nice. And that factor of $2$ at the top end gets killed off pretty early in the race between $n^{2n} $ and $2n!$ - by $n=3$ in fact, even though the $(1-\frac{2^2}{3^2})$ term isn't quite enough to cancel it out on its own. – Joffan Jun 07 '16 at 06:00
  • yes it does , but to remove any doubt i multiplied it with the smallest term possible :P – avz2611 Jun 07 '16 at 06:01
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Lemma : For $k=0,1,\cdots,199$, $$(600-2k)(599-2k)(k+1)\lt 300^3$$ Using the lemma, $$\begin{align}600!&=\prod_{k=0}^{199}(600-2k)(599-2k)(k+1)\lt (300^3)^{200}=3^{600}\end{align}$$

Proof for lemma :

$$(600-2x)(599-2x)(x+1)\lt 300^3$$ is equivalent to $$4 x^3-2394 x^2+357002 x-26640600\lt 0$$ Let $f(x)$ be the LHS. Then, $$f'(x)=12 x^2-4788x+357002$$ Let $x=\alpha$ be the smaller solution of $f'(x)=0$. Then, noting that $\alpha\gt 0$, we have $$f(\alpha)=\frac{2i-399}{6}f'(\alpha)-\frac{241202}{3}\alpha-2899967\lt 0$$ With $f(199)\lt 0$, we have $f(x)\lt 0$ for $0\le x\le 199$. $\blacksquare$

mathlove
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  • Great method, I wish you used a different symbol instead of $i$ – ar2015 Jun 07 '16 at 05:35
  • @ar2015: OK. I edited. Thank you. – mathlove Jun 07 '16 at 05:37
  • you replaced some $i$ and left some others. BTW, $i$ reminds $\sqrt{-1}$ and $x, y$ remind real values. $k, m, n, p, q$ remind integer numbers. Despite there is no restriction and any symbol you use is ok. – ar2015 Jun 07 '16 at 05:48
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$$\ln{n!} = \sum_{k=1}^n\ln{k}$$ solve for $n!$ : $$n! = e^{\sum_{k=1}^n\ln{k}}$$ approximate $ \sum_{k=1}^{600}\ln{k} \approx 3242.2753...$ $$600!\approx e^{3242.2753...}$$ On the other hand we can express $300^{600}$ as $e^{\ln{(300^{600})}}$. Then using logarithm rules: $$300^{600}=e^{600\ln{300}}\approx e^{3422.2694...}$$ $$e^{3242.2753...}<e^{3422.2694...}$$ $$600! < 300^{600}$$ (P.S. This is one of my first posts, I'm sorry if it's not the best work.)

Noah Geren
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