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**Sampras** 1. Prove that $\displaystyle n! > \left(\frac{n}{e} \right)^{n} $ for $\displaystyle n \geq 1 $.

*Proof*. We use induction on $\displaystyle n $. For $\displaystyle n=1 $, $\displaystyle 1 > \frac{1}{e} $. Now suppose that for some $\displaystyle k > 1 $, $\displaystyle k! > \left(\frac{k}{e} \right)^{k} $. So we want to show that $\displaystyle (k+1)! > \left(\frac{k+1}{e} \right)^{k+1} $. Now $\displaystyle (k+1)! > (k+1) \cdot \left(\frac{k}{e} \right)^{k} $. Can we somehow use the fact that $\displaystyle \left(1+ \frac{1}{n}\right)^{n} < e $ for all $\displaystyle n $? We could flip it so that $\displaystyle \frac{1}{e} < \frac{1}{\left(1+\frac{1}{n} \right)^{n}} $ for all $\displaystyle n $. And so $\displaystyle \left(\frac{n}{e} \right)^{n} < \frac{n}{\left(1+\frac{1}{n} \right)^{n^2}} $. Or another approach may be to show that $\displaystyle \frac{k^{k} \cdot (k+1)}{e^k} > \frac{(k+1)^{k+1}}{e^{k+1}} $? The denominator of the second term is larger, indicating that it would probably be smaller.

2. Show that $\displaystyle n! < \left(\frac{n+1}{n} \right)^{n} $ for $\displaystyle n>1 $, and deduce that $\displaystyle n! < e \left(\frac{n}{2} \right)^{n} $.

For this one you would use the AM-GM inequality? Because the LHS is $\displaystyle n(n-1)(n-2) \cdots 1 $. So taking $\displaystyle \sqrt[n]{n!} $ we can use the AM-GM inequality?