4

How do I prove this by induction:

$$\displaystyle n^{n+1} > (n+1)^n,\; \mbox{ for } n\geq 3$$ Thanks.

What I'm doing is bunch of these induction problems for my first year math studies.

I tried using Bernoulli's inequality at some point, but no success. Also, tried $(n+1)^{n+2}=(n+1)^{n+1}(n+1)$, then expanding $(n+1)^{n+1}$ by binomial formula to get the $n^{n+1}$ member to apply the induction hypothesis, still no success.

Here's another one I've been struggling with:

$$(n!)^2 \leq \left(\frac{(n + 1)(2n + 1)}{6}\right)^n$$

EDIT: Finally solved the second one!

What I needed was the AM-GM inequality.

Therefore,

$$\frac{(n + 1)(2n + 1)}{6} = \frac{1}{n} \sum_{i=1}^{n} i^2 \geq \sqrt[n]{1^2 \cdot 2^2 \cdots n^2}$$

Thus,

$$\left(\frac{(n + 1)(2n + 1)}{6}\right)^n = \left(\frac{1}{n} \sum_{i=1}^{n} i^2\right)^n \geq 1^2 \cdot 2^2 \cdots n^2 = (n!)^2$$

Done.

Martin Sleziak
  • 53,687
eudoxyz
  • 128

5 Answers5

6

it is easier if you rearrange the statement a little. dividing throughout by the +ve quantity $n^n$ it becomes the claim that

$$ \left(1+\frac1n\right)^n \lt n $$ which is true if $n=3$.

now multiply both sides by the +ve quantity $\left(1+\frac1n\right)$.

robjohn
  • 345,667
David Holden
  • 18,040
3

HINT for the induction step: If your induction hypothesis is that $n^{n+1}>(n+1)^n$, then

$$(n+1)^{n+2}=(n+1)n^{n+1}\left(\frac{n+1}n\right)^{n+1}>(n+1)^{n+1}\left(\frac{n+1}n\right)^{n+1}\;.\tag{1}$$

Now combine everything on the righthand end of $(1)$ into a single $(n+1)$-st power and do a little algebra.

Brian M. Scott
  • 616,228
3

Your inequality is the same as $$ n\left(\frac{n}{n+1}\right)^n\gt1 $$ Notice that $$ \begin{align} (n+1)\left(\frac{n+1}{n+2}\right)^{n+1} &=\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+1}n\left(\frac{n}{n+1}\right)^n\\ &=\left(1+\frac1{n(n+2)}\right)^{n+1}n\left(\frac{n}{n+1}\right)^n\\[9pt] &\gt1\cdot1 \end{align} $$

robjohn
  • 345,667
3

Supposed for $n\geq 3$ you have that $n^{n+1}>(n+1)^n$. WTS $(n+1)^{n+2}>(n+2)^{n+1}$. Since $n^{n+1}>(n+1)^n$ you get that $n^{n+1} \cdot \frac{(n+1)^{n+2}}{n^{n+1}}>(n+1)^n\cdot \frac{(n+1)^{n+2}}{n^{n+1}}=\frac{(n+1)^{2n+2}}{n^{n+1}}>(n+2)^{n+1}$, where the last inequality follows since $(n+1)^2>n(n+2)$. Hence $(n+1)^{n+2}>(n+2)^{n+1}$.

mathemagician
  • 3,225
1

Assume the statement holds true for $n-1$, i.e. we have $$(n-1)^n>n^{n-1}.$$

Then \begin{align*} (n+1)^n &= n^n+\binom{n}{1}n^{n-1}+\dots+\binom{n}{n-1}n+1\\ &<n^n+n^n+\dots+n^n=n(n^n)=n^{n+1}. \end{align*}

BlackAdder
  • 4,029