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Which is larger $\sqrt[99]{99!}$ or $\sqrt[100]{100!}$

I know that it is the $\sqrt[100]{100!}$

but is there a formula to figure this out instead of doing it all out by hand?

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18  
Please do not do this out by hand. That isn't good for your health. :) –  NotNotLogical Apr 21 at 17:25
23  
...or your hand –  DepeHb Apr 21 at 17:27

6 Answers 6

Notice that $\sqrt[99]{99!}$ is the geometric mean of all the integers from 1 to 99, while $\sqrt[100]{100!}$ is the geometric mean of those same integers and 100. What happens to a mean when you include one more element that is larger than all the previous ones?

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Isn't this a bit circular? Surely it's this fact that is why the geometric mean is useful, and is called a mean. –  LTS May 21 at 18:35

If you bring both to the $99*100$ power, you get the two numbers$$(99!)^{100},(100!)^{99}=(99!)^{99}100^{99}$$ So dividing out the common factor we want to compare $$99!,100^{99}$$ It should be clear which is larger.

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Let $x=\sqrt[99]{99!}$ and $y=\sqrt[100]{100!}$, then \begin{align} \frac{x}{y}&=\frac{\sqrt[99]{99!}}{\sqrt[100]{100!}}\\ &=\frac{(99!)^\frac{1}{99}}{(100!)^\frac{1}{100}}\\ &=\frac{(99!)^{\frac{1}{100}+\frac{1}{9900}}}{(100!)^\frac{1}{100}}\\ &=\frac{(99!)^{\frac{1}{100}}(99!)^{\frac{1}{9900}}}{(100!)^\frac{1}{100}}\\ &=(99!)^{\frac{1}{9900}}\left(\frac{99!}{100!}\right)^\frac{1}{100}\\ &=(99!)^{\frac{1}{9900}}\left(\frac{99!}{99!\cdot100}\right)^\frac{1}{100}\\ &=\frac{(99!)^{\frac{1}{9900}}}{100^\frac{1}{100}}\\ \left(\frac{x}{y}\right)^{9900}&=\left(\frac{(99!)^{\frac{1}{9900}}}{100^\frac{1}{100}}\right)^{9900}\\ &=\frac{99!}{100^{99}}\\ &=\frac{99\cdot98\cdot97\cdots1}{100\cdot100\cdot100\cdots100}\\ \frac{x}{y}&=\sqrt[9900]{\frac{99\cdot98\cdot97\cdots1}{100\cdot100\cdot100\cdots100}} \end{align} It is clearly $y>x$, hence $\sqrt[100]{100!}>\sqrt[99]{99!}$.

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The logarithm of $\sqrt[99]{99!}$ is $\frac{1}{99}\log 99!=\frac{1}{99}\sum_{i=1}^{99}\log i$. That is, it's the average of the logarithms of integers from $1$ to $99$. Similarly, the logarithm of $\sqrt[100]{100!}$ is the average of the logarithms of integers from $1$ to $100$. Clearly the latter average is larger; so $\sqrt[100]{100!}$ is larger too.

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Really like this answer. –  DepeHb Apr 21 at 23:19

Hint: Let $x=\sqrt[99]{99!}$ and $y=\sqrt[100]{100!}$. Then $x^{99}=99!$ and $y^{100}=100!=100\cdot 99!=100\cdot x^{99}$. What is an upper bound on $x$?

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I guess I'm too late, and it's not so sleek, but I couldn't restrain myself.

So on LHS you have $(99!)^{\frac{1}{99}}$, on RHS $(100!)^{\frac{1}{100}}$. By logging both sides you get $$ \frac{\log 99!}{99} \ \ \ \frac{\log 100!}{100}\\ \frac{\sum_{k=1}^{99}\log k}{99} \ \ \ \frac{\sum_{k=1}^{100}\log k}{100}\\ 100 \sum_{k=1}^{99}\log k \ \ \ 99\sum_{k=1}^{100}\log k\\ $$ Rewriting the last expression as $99 \cdot \sum_{k=1}^{99}\log k + 99 \log 100$, subtract $99 \cdot \sum_{k=1}^{99}\log k$ on both sides, and since the number of terms in both sums is the same, on LSH you get $\sum_{k=1}^{99} \log k = \log 99!$ and on RHS $99 \log 100 = \log 100^{99}$. Now exponentiate both sides, on LHS you have $1 \cdot 2 \cdots 99$, and on RHS $100 \cdot 100 \cdots 100$, and the number of terms is the same.

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