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Another question regarding spheres:

Is it true that $|S_{n}| = O( \frac{1}{\sqrt n} )$ ?

(I mean the surface area of the n dimensional sphere) Thanks !

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    You've seen [this](http://mathworld.wolfram.com/Hypersphere.html)?2012-07-23
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    I did... I know the formula for the surface area of the sphere, but just wondering if this gamma function can be approximated in a way that will give us that the surface area is $O(\frac{1}{\sqrt n } ) $ . Hope I formulated it in a clearer way now.2012-07-23

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Use Stirling's approximation $\Gamma(\frac n2)=(\frac n2 -1)!\approx (n/2)^{n/2}e^{-n/2}$, so $$ S_n=\frac{2\pi^{n/2}}{\Gamma(\frac12 n)} \approx\frac{2\pi^{n/2}}{(\frac n2-1)^{\frac n2-1}e^{-\frac n2+1}} =2\left(\frac{\pi e^{1-2/n}}{(n/2-1)^{1-2/n}}\right)^{n/2}\\ \approx \left(\frac1n\right)^{n/2} =\left(\frac1{\sqrt{n}} \right)^{n} < \left(\frac1{\sqrt{n}} \right), $$ for $n>1$ (then $(n/2-1)^{1-2/n}\sim (n/2)\;$ for large $n$). So $S_n$ doesn't grow faster than $O\left(\frac1{\sqrt{n}}\right)$.

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    Since $n^n/n!\to\infty$, I have some trouble following your estimations.2012-07-23
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    @MartinArgerami, would you feel less troubled, if I write $n!\approx n^ne^{-n}$? Then $(n^ne^{-n}/n!\to 0)$ and an additional $e$ would appear: $\left(\frac{2\sqrt{n}e}{n/2} \right)^{n/2}$, but the rest would stay the same...right?2012-07-23
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    I would still feel troubled, because you are replacing the value of the Gamma function with $n!$, its value at integer points. But you want $\Gamma(n/2)$, so it would probably be right to usema direct estimate for the Gamma function (which is the one you mention). Also, right now the terms on the sides of your second equal sign, are not equal.2012-07-23
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    @MartinArgerami Thanks for spotting that, I edited it...better now?2012-07-23
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    Better... But still, you are using the approximation for $\Gamma(n/2+1)$ and not for $\Gamma(n/2)$. And in your first term in brackets, the $2$ should outside the brackets, and $\sqrt\pi$ should be $\pi$.2012-07-23
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    @draks: Thanks a lot! When you're writing $\approx $ , it means that one is big-O of another? So actually, we do have a constant in the RHS (if we 're using the big-O notation) ? Anyway, like MartinArgerami said, you're using the approximation of $\Gamma(n/2 +1 ) $ ... Is it "fixable"? Thanks again !2012-07-23
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    @joshua I meant $\approx$ just as an approximation. My overall idea was to show that $S_n$ shrinks much faster than $O(1/\sqrt{n})$ and I thought I fix my $\Gamma$ mistake. All-in-all you're welcome and I'm happy that I could help...2012-07-23