1. ## Inequalities

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

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

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

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

2. Originally Posted by Sampras
1. Prove that $n! > \left(\frac{n}{e} \right)^{n}$ for $n \geq 1$.

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

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

For this one you would use the AM-GM inequality? Because the LHS is $n(n-1)(n-2) \cdots 1$. So taking $\sqrt[n]{n!}$ we can use the AM-GM inequality?
For (1), $k^{k} \cdot (k+1) \not > (k+1)^{k+1}$. So the inequality depends on the denominators $e^{k}$ and $e^{k+1}$. Is this where we use the fact that $\left(1+\frac{1}{n} \right)^{n} < e$ for all $n$?

3. Originally Posted by Sampras
1. Prove that $n! > \left(\frac{n}{e} \right)^{n}$ for $n \geq 1$.

Proof. We use induction on $n$. For $n=1$, $1 > \frac{1}{e}$. Now suppose that for some $k > 1$, $k! > \left(\frac{k}{e} \right)^{k}$. So we want to show that $(k+1)! > \left(\frac{k+1}{e} \right)^{k+1}$. Now $(k+1)! > (k+1) \cdot \left(\frac{k}{e} \right)^{k}$.
so we just need to prove that $(k+1)\left(\frac{k}{e} \right)^k > \left(\frac{k+1}{e} \right)^{k+1},$ which after simplifying gives us this equivalent inequality: $\left(1 + \frac{1}{k} \right)^k < e,$ which we know is true.

2. Show that $n! < \left(\frac{n+1}{n} \right)^{n}$ for $n>1.$
this is false! check what you wrote again!

4. Originally Posted by NonCommAlg
so we just need to prove that $(k+1)\left(\frac{k}{e} \right)^k > \left(\frac{k+1}{e} \right)^{k+1},$ which after simplifying gives us this equivalent inequality: $\left(1 + \frac{1}{k} \right)^k < e,$ which we know is true.

this is false! check what you wrote again!
It should be $n! < \left(\frac{n+1}{2} \right)^{n}$. So basically you simplified the given inequality into an inequality we know?

5. Originally Posted by Sampras
It should be $n! < \left(\frac{n+1}{2} \right)^{n}$.
this is a very straightforward application of AM-GM: $\frac{n+1}{2}=\frac{\frac{n(n+1)}{2}}{n}=\frac{1+2 + \cdots + n}{n} > \sqrt[n]{1 \cdot 2 \cdot \cdots n}=\sqrt[n]{n!}.$

6. Originally Posted by NonCommAlg
this is a very straightforward application of AM-GM: $\frac{n+1}{2}=\frac{\frac{n(n+1)}{2}}{n}=\frac{1+2 + \cdots + n}{n} > \sqrt[n]{1 \cdot 2 \cdot \cdots n}=\sqrt[n]{n!}.$
And to deduce that $n! < e \left(\frac{n}{2} \right)^{n}$ we know that $n! < \left(\frac{n+1}{2} \right)^{n}$. So maybe simplify the first inequality to get the second one? We can do the following: $n! < \left[\left(1+\frac{1}{n} \right) \cdot \frac{n}{2} \right]^{n}$? We then get the second inequality.

Is this correct?

7. Originally Posted by Sampras
And to deduce that $n! < e \left(\frac{n}{2} \right)^{n}$ we know that $n! < \left(\frac{n+1}{2} \right)^{n}$. So maybe simplify the first inequality to get the second one? We can do the following: $n! < \left[\left(1+\frac{1}{n} \right) \cdot \frac{n}{2} \right]^{n}$? We then get the second inequality.

Is this correct?
you just need to show that $\left(\frac{n+1}{2} \right)^n < e \left(\frac{n}{2} \right)^n,$ which, again, after simplifying gives you this equivalent and true inequality: $\left( 1+ \frac{1}{n} \right)^n < e.$

8. Originally Posted by NonCommAlg
you just need to show that $\left(\frac{n+1}{2} \right)^n < e \left(\frac{n}{2} \right)^n,$ which, again, after simplifying gives you this equivalent and true inequality: $\left( 1+ \frac{1}{n} \right)^n < e.$

But my simplification of the current inequality to the inequality we proved is incorrect?