1

I got some question on how to proceed on the proof below,

Prove that:

$n^{n+1}>(n+1)^{n}$, for $n\geq3$

By induction:

Inequality holds for $n=3$ , $3^4=81\geq 4^3 =64$.

Suppose it holds for $k^{k+1}>(k+1)^{k}$.

Prove for $k+1$ :

$(k+1)^{k+2}\geq(k+2)^{k+1}$

and here is the part where I am kind of stuck, how would I use the hypothesis to prove for $k+1$?

Cookie
  • 13,532
evaz34
  • 25
  • 2
  • I see a "$k+1$" in your inductive hypothesis and your last inequality. What can you do to make the one occurrence exactly like the other? – Eric Towers Jun 15 '14 at 23:06
  • Are you required to use induction? If not, you can instead just show that $\dfrac{\ln x}{x}$ is strictly decreasing for $x \ge 3$. From which, it follows that $\dfrac{\ln(n+1)}{n+1} < \dfrac{\ln n}{n}$, i.e. $(n+1)^n < n^{n+1}$. – JimmyK4542 Jun 15 '14 at 23:10

1 Answers1

7

Your inequality is equivalent to

$$n> \left(1+\frac1n\right)^n.$$

Assuming it holds for some $n\ge 3$ we have to show it for $n+1.$ We have:

$$\left(1+\frac1{n+1}\right)^{n+1}< \left(1+\frac1n\right)^{n+1}=\left(1+\frac1n\right)\left(1+\frac1n\right)^n< \left(1+\frac1n\right)n=n+1,$$ what finishes the proof.

mfl
  • 29,399