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Which is greater among $300 !$ and $\sqrt {300^{300}}$ ?

The answer is $300 !$ (my textbook's answer). I do not know how to solve problems involving such large numbers.

Michael Lugo
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Apurv
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    Simplify the second term a little bit (in other words, write $(300^{300})^{1/2}$ as $300^{150}$. Now, expand the terms. – muffle Dec 12 '13 at 07:07
  • See http://math.stackexchange.com/questions/46892/how-do-you-prove-that-nn-is-on2. – lhf Dec 12 '13 at 13:54

3 Answers3

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HINT: $(300^{300})^{1/2}=300^{150}$, so you’re comparing

$$\underbrace{300\cdot300\cdot\ldots\cdot300}_{150\text{ factors}}$$

with

$$300\cdot299\cdot\ldots\cdot1=\underbrace{(300\cdot1)\cdot(299\cdot2)\cdot\ldots\cdot(151\cdot150)}_{150\text{ factors}}\;.\tag{1}$$

Show that each of the parenthesized factors in $(1)$ is at least $300$.

Brian M. Scott
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When calculations involve large factorials, the use of Stirling approximation for $n!$ is very convenient. In your case, you compare $n!$ to $n^{n/2}$. We can take logarithms of both sides and use the fact that Stirling approximation gives $\log(n!)$ close to $n \log(n) - n$; the logarithm of the second term is simply $\dfrac{n \log(n)} {2}$. Using this approximation, you find that the factorial is the largest as soon as $n > 8$. Without any approximation, the factorial is the largest as soon as $n > 2$ [$n=2$ is the only solution of $n! = n^{n/2}$]

Apurv
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Claude Leibovici
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$300!=(300*1) * (299*2) * ... * (151*150)$, 150 pair products total

$(300^{300})^{1/2}= 300^{150} = 300 * 300 * ... * 300$, 150 counts of 300 total

none of pair product in the first line is smaller than $300$, so $300!$ is larger.

MoonKnight
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