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I was deliberately trying to solve this problem for the last 2 hours. I tried solving it using substitution, different notations, etc. but that did not lead me to an answer.

Could someone please help me?

Thanks in advance!

Thomas Andrews
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Flower
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  • Functionally, there is no difference between $f(n)=O(g(n))$ and $f(n)\leq O(g(n)).$ We usually don’t write $\leq$ for that reason. – Thomas Andrews Oct 22 '21 at 22:17
  • I’d use Stirling’s approximation, if you know that result. https://en.wikipedia.org/wiki/Stirling%27s_approximation – Thomas Andrews Oct 22 '21 at 22:20
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    But I’m not sure it is even true. Stirling say $$n!\sim (n/e)^n\sqrt{2\pi n}.$$ Seems like $e^{n}$ is a lot less than $n^{n/2}$ when $n$ is large. Specifically: $$\frac{e^n}{n^{n/2}}\to 0.$$ – Thomas Andrews Oct 22 '21 at 22:26
  • Thank you very much! I didn’t know about Stirling‘s approximation but that helped! Thanks again! – Flower Oct 23 '21 at 12:10

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