Proving $k^k>(k+1)^{k-1}$

Question

For $k>2,$ Show $k^k<(k!)^2$

Using induction, I boiled it down to proving $k^k>(k+1)^{k-1}$

Im stuck here. I tried using binomial inequality that i.e. $$k^k=((k+1)-1)^k>(k+1)^k-k(k+1)^{k-1}=(k+1)^{k-1}$$

But the fact that it has a negative number may not be the right argument, so, how to proceed?

Answer 1

HINT

We have that

$$k^k>(k+1)^{k-1}\iff k\log k>(k-1)\log(k+1)\iff\frac{\log k}{k-1}>\frac{\log(k+1)}{k}$$

then consider $f(x)=\frac{\log x}x\implies f'(x)=\frac{1-\log x}{x^2}<0$ as $x>e$.

Answer 2

Note that $\frac{(k+1)!)^2}{k!^2}=(k+1)^2$ and that$$\frac{(k+1)^{k+1}}{k^k}=(k+1)\left(1+\frac1k\right)^k.$$Now, use the fact that$$(k+1)\left(1+\frac1k\right)^k<(k+1)\times e$$in order to prove that $\frac{(k+1)^{k+1}}{k^k}<\frac{(k+1)!)^2}{k!^2}$.

Answer 3

Without using calculus:

• You can write $$(k!)^2=\prod_{n=1}^{k}n(k-n+1)$$ that is, $(k!)^2=(1\cdot k)\times (2\cdot (k-1))\times \ldots \times (k\cdot 1)$.

• Show that $n(k-n+1)\geq k$. This is the same as showing that, when $1\leq n\leq k$, $$n^2-n(k+1)+k=(n-1)(n-k)\leq 0$$ which is easily seen to be true since $n-k$ is always negative or $0$, and $n-1$ is always positive or $0$. Notice that $(n-1)(n-k)<0$ (strict inequality) if $n\neq1$ and $n\neq k$. Because of this, we get our strict inequality precisely when there is atleast one $n$ such that $1<n<k$ (which fails when $k=2$).

Thus, when $k>2$, $$(k!)^2=\prod_{n=1}^{k}n(k-n+1)> \prod_{n=1}^{k}k=k^k$$

Answer 4

Here's another non-calculus demonstration. Given $k \geq 2$, we first establish that for $1 \leq j < k$, $\left(1+\frac{1}{k}\right)^j < j+1$.

This is evidently true for $j = 1$:

$$ \left(1+\frac{1}{k}\right)^1 < 2 = 1+1 $$

since $k \geq 2$. Now suppose it to be true for $j-1$, with $1 < j < k$:

$$ \left(1+\frac{1}{k}\right)^{j-1} < j $$

Then the claim follows by induction:

$$ \left(1+\frac{1}{k}\right)^j < j\left(1+\frac{1}{k}\right) = j + \frac{j}{k} < j+1 $$

since $j < k$.

In particular, let $j = k-1$, in which we get

$$ \left(1+\frac{1}{k}\right)^{k-1} < k $$

Multiply both sides by $k^{k-1}$ and we get

$$ (k+1)^{k-1} < k^k $$

as desired.

Answer 5

The easiest proof is to invoke the AM-GM inequality: we have $$ \sqrt[k]{(k+1)^{k-1}} =\sqrt[k]{(k+1)^{k-1}\times1} \stackrel{GM\le AM}{\le}\frac{(k-1)(k+1)+1}{k} =k $$ but strict inequality must hold because we are taking the geometric mean of $k$ numbers that are not all equal to each other.