Show by induction that $n^n>2^{n+1}$

Question

I have been asked to prove $n^n>2^{n+1}$ for every natural number $\ge$ 3

I have understood the Induction Basis and Hypothesis :

Induction Basis : n=3 ; 27>16

Induction Hypothesis : let n $\ge$ 3 such that $n^n>2^{n+1}$

However, I struggle to understand the following statement

Induction Step: $(n + 1)^{n+1} > n^n· n >(IH) 2^{n+1}· n > 2 ^{n+2}$

My issue with the IS is how one arrives at $2^{n+1}· n$ from $2 ^{n+2}$.

I can see that with the LHS of the IS has been expanded and any constants have been removed because they are of 'lower impact' than the other $n^n$ expression but can't see how the issue above works.

Does one really need induction? It's clear for $n=3$. For $n\ge 4$, we have $n^n=n \cdots n \cdot n \ge 2 \cdots 2 \cdot 4 = 2^{n-1} \cdot 4 = 2^{n+1}$. — lhf, Dec 15 '20 at 12:08

score 3 · Answer 1 · answered Dec 28 '19 at 12:18

3

Since $n\geqslant3$, then $n>2$, and therefore $2^n\times n>2^n\times2=2^{n+1}$.

answered Dec 28 '19 at 12:18

José Carlos Santos

427,504

I am getting confused by this final step, the induction step, and am struggling to understand these responses in relation to the question. Are you able to help me understand the process behind this step and what I should be looking for? – Unknown Dec 28 '19 at 14:27
It is not clear to me what your question is. I simply used the fact that$$n>2\implies2^n\times n>2^n\times2.$$ – José Carlos Santos Dec 28 '19 at 14:28
I can understand why, I'm not explaining myself well enough. So the idea with the induction hypothesis is to assume that the statement in the question is correct. Then in the Induction step we, using the hypothesis assumption, apply that to the $(n + 1)$ which allows us to prove the question. Am I right? – Unknown Dec 28 '19 at 14:35
If we have a sequence $P(k),P(k+1),P(k+2),\ldots$ of propositions, proving by induction that they all hold means proving that $P(k)$ holds and that if, for some $n\in{k,k+1,k+2,\ldots}$, $P(n)$ holds, then $P(n+1)$ holds too. So, in your case, proving the the $(\forall n\geqslant 3):n^n>2^{n+1}$ means to prove that $3^3>2^4$ (and you did that) and to prove that if, for some $n\in{3,4,5,\ldots}$, $n^n>2^{n+1}$, then $(n+1)^{n+1}>2^{n+2}$. – José Carlos Santos Dec 28 '19 at 14:39
Reference : 'and to prove that if, for some $n∈{3,4,5,…}, nn>2n+1$' is this the worded version of an induction hypothesis? – Unknown Dec 28 '19 at 14:54
Yes, it is. ${}$ – José Carlos Santos Dec 28 '19 at 15:15

Calvin Khor · Accepted Answer · 2019-12-29T01:43:06.867

2

I can see that with the LHS of the IS has been expanded and any constants have been removed because they are of 'lower impact' than the other $n^n$ expression but can't see how the issue above works.

I don't see what has been expanded and where any constants have been removed.

The statement that we need to prove is:

Induction step. If $n\ge 3$ and $n^n > 2^{n+1}$, then $(n+1)^{n+1} > 2^{n+2}$.

To do this: first, for any integers $x>0$ and $n>0$, it is elementary that $(x+1)^{n+1} > x^{n+1}$. Hence, for $x = n$,

$$(n+1)^{n+1} > n^{n+1} = \color{blue}{n^n} \cdot \color{red}n.$$

Now, the induction hypothesis is that $\color{blue}{n^n > 2^{n+1}}$. So we plug this in to get $$ \color{blue}{n^n} \cdot {\color{blue}{\color{red}n > 2^{n+1}}} \cdot \color{red}n.$$ And we haven't yet used the assumption $\color{red}{n\ge 3}$; we should do so now (as the comment of Eureka points out) $$ \color{blue}{2^{n+1}} \cdot \color{red}{n \ge \color{blue}{2^{n+1}} \cdot 3} > 2^{n+1} \cdot 2 = 2^{n+2}.$$ This completes the induction step, by joining the three inequalities together.

edited Dec 29 '19 at 01:43

answered Dec 28 '19 at 12:39

Calvin Khor

34,903

I am getting confused by this final step, the induction step, and am struggling to understand these responses in relation to the question. Are you able to help me understand the process behind this step and what I should be looking for? Always struggle with understanding this maths – Unknown Dec 28 '19 at 14:27
@Richard I tried to make it clearer. I had a small mistake earlier, but its correct now. Re: your comment, (1) My entire response is only about the induction step. (2) Its hard for me to say, because you're the one who doesn't understand :) Can you try to point out which step or which inequality you don't understand? Or do you not understand the whole point of the induction step? – Calvin Khor Dec 29 '19 at 01:44
@Richard , looking at your comments under José's answer, I think that you don't need help with this question specifically, or the induction step of this question; I think you don't understand induction itself. For this, you could try to read say, https://math.stackexchange.com/questions/19485/dominoes-and-induction-or-how-does-induction-work ... if you've more questions I'd be happy to discuss – Calvin Khor Dec 29 '19 at 01:52

score 1 · Answer 3 · answered Dec 28 '19 at 14:11

1

If your formula is exact, the induction hypothesis is $\;n^n>2^{n\color{red}{+1}}\enspace (n\ge 3)$. So \begin{align} (n+1)^{n+1}=(n+1)^n(n+1)>n^n\cdot n &>2^{n+1}\cdot 2=2^{n+2}.\cr &\downarrow \\ \text{(inductive}&\text{ hypothesis)} \end{align}

answered Dec 28 '19 at 14:11

Bernard

175,478

I am getting confused by this final step, the induction step, and am struggling to understand these responses in relation to the question. Are you able to help me understand the process behind this step and what I should be looking for? – Unknown Dec 28 '19 at 14:26
I thought have made it as clear as possible. The other answers seem to understand that what should be proved is $n^n>2^n$, and I supposed your formula was correct, i.e. it is $>2{n+1}$, so that the formula that has to be proved in this inductive step is $;(n+1)^{n+1}>2^{n+1}$. Is it clearer? – Bernard Dec 28 '19 at 14:32

Show by induction that $n^n>2^{n+1}$

3 Answers3