Induction proof of $n^{(n+1) }> n(n+1)^{(n-1)}$

Question

The question statement from my homework booklet goes:

Prove by mathematical induction that $n^{n+1} > n(n+1)^{n-1}$ is true for all integers $n \geq 2$.

I've managed to come up with this for the induction step (the base case is trivial), but I am not sure what to do from here: Assume true for n=k. For n=k+1, \begin{align*} &k^{k+1} > k(k+1)^{k+1} \\ &(k+1)k^{k+1} > k(k+1)^k \\ &(k+1)^{k+2} < (k+1)^3 k^k < (k+1)^3 (k+2)^k \\ &(k+1)^{k-1} < (k+2)^k \end{align*} I would greatly appreciate any help with how to solve this. Thanks in advance.

Answer 1

At step 3: when n = k + 1 the inequality becomes: (k+1)^(k+2) > (k+1)*(k+2)^k <===>

(k+1)^(k+1) > (k+2)^k <===> (k+1)^(k+1) > ((k+1) + 1)^k <===> k+1 > (1 + 1/(k+1))^k ) (*).

(*) is true because:

1. k+1 > 3

2. 3 > (1 + 1/(k+1))^(k+1) ( reminder: the limit of the right side is e = 2.71728 < 3 )

3. (1 + 1/(k+1))^(k+1) > (1 + 1/(k+1))^k

Answer 2

We are required to prove that: $(k+1)^{k+2}>(k+1)(k+2)^k$.

The assumption is that: $k^{k+1} > k(k+1)^{k-1}$. Dividng both sides by $k^k$ gives: $k> \left(\frac{k+1}{k}\right)^{k-1}$, thus $k>\left(1+\frac{1}{k}\right)^{k-1}$

Starting from this, \begin{eqnarray} k&>&\left(1+\frac{1}{k}\right)^{k-1}\\ \frac{1}{\left(1+\frac{1}{k}\right)^{k-1}}&>&\frac{1}{k}\\ 1+\frac{1}{\left(1+\frac{1}{k}\right)^{k-1}}&>&1+\frac{1}{k}\\ \left(1+\frac{1}{k}\right)^{k-1}+1&>&\left(1+\frac{1}{k}\right)^k\\ \end{eqnarray}

Now, since $k>\left(1+\frac{1}{k}\right)^{k-1}$, then: \begin{eqnarray} k+1&>&\left(1+\frac{1}{k}\right)^k\\ (k+1)^{k+2}&>&(k+1)^{k+1}\left(1+\frac{1}{k}\right)^k \end{eqnarray}

Also, $\frac{1}{k}>\frac{1}{k+1}$, therefore: \begin{eqnarray} (k+1)^{k+2}&>&(k+1)^{k+1}\left(1+\frac{1}{k+1}\right)^k\\ &>&(k+1)(k+2)^k \end{eqnarray} Therefore, $(k+1)^{k+2}>(k+1)(k+2)^k$. $\Box$

Answer 3

Define: $\rm f(n) = \dfrac{n^{n+1}}{n(n+1)^{n-1}} \implies \dfrac{f(n+1)}{f(n)} = \dfrac{(n+1)^{2n}}{n^n(n+2)^n} = \left(\dfrac{n^2 + 2n + 1}{n^2 + 2n}\right)^n > 1$

Base Case: $\rm f(2) = \dfrac{2^3}{2 \cdot 3^1} = \dfrac{8}{6} > 1$

Inductive Step: $\rm f(n) > 1 \implies f(n+1) = \underbrace{\left(\dfrac{f(n+1)}{f(n)}\right)}{} \cdot \underbrace{f(n)}{} > 1$ since both terms in the product are greater than $1$ individually.

See here and here for other examples of induction using multiplicative telescopy, a very powerful technique illuminated by Bill Dubuque.

Answer 4

For $k$ we have

$\large k^{k+1}>k(k+1)^{k-1}$

which we hold to be true.

For $k+1$ we have

$\large(k+1)^{k+2}=(k+1)^{k+1}(k+1)=k^{k+1}(1+\frac{1}{k})^{k+1}(k+1) \\\large>k(1+k)^k(1+\frac{1}{k})^{k+1}$

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

Now

$\large k(k+1)^k(1+\frac{1}{k})^{k+1}=k(k+1)^k(1+\frac{1}{k})^{k+1}(\frac{k+2}{k+2})^k \\\large =k(k+1)^k(\frac{k+1}{k})^{k+1}(\frac{k+2}{k+2})^k=(\frac{(k+1)^2}{k(k+2)})^k(k+1)(k+2)^k\\\large>(k+1)(k+2)^k$

where we have used the fact that $(k+1)^2=k(k+2)+1>k(k+2)$

so that $\large (\frac{(k+1)^2}{k(k+2)})^k>\frac{(k+1)^2}{k(k+2)}>1$

Thus we have

$\large (k+1)^{k+2}>k(1+k)^k(1+\frac{1}{k})^{k+1}>(k+1)(k+2)^k$

Culminating in

$\large (k+1)^{k+2}>(k+1)(k+2)^k$