Proving by induction: $2^n > n^3 $ for any natural number $n > 9$

Question

I need to prove that $$ 2^n > n^3\quad \forall n\in \mathbb N, \;n>9.$$

Now that is actually very easy if we prove it for real numbers using calculus. But I need a proof that uses mathematical induction.

I tried the problem for a long time, but got stuck at one step - I have to prove that:

$$ k^3 > 3k^2 + 3k + 1 $$

Hints???

$k \leqslant k^2$, $1 \leqslant k^2$. – Daniel Fischer Jul 07 '13 at 17:20 — Daniel Fischer, Jul 07 '13 at 17:20

score 13 · Answer 1 · answered Jul 07 '13 at 19:09

For another way just using $n>9$, note that when $n=10$, $2^n = 1024 > 1000 = n^3$. Now suppose that $2^n>n^3$ for $n>9$. Then,

$\begin{align*} 2^{n+1} &= 2\cdot2^n \\ &>2n^3 \\ &= n^3 +n^3 \\ &> n^3 + 9n^2 \\ &= n^3 + 3n^2 + 6n^2 \\ &>n^3 + 3n^2 +54n \\ &=n^3+3n^2+3n +51n\\ &>n^3+3n^2+3n+1 \\ &= (n+1)^3. \end{align*}$

amWhy · Accepted Answer · 2013-07-07T17:31:03.670

5

For your "subproof":

Try proof by induction (another induction!) for $k \geq 7$

$$k^3 > 3k^2 + 3k + 1$$

And you may find it useful to note that $k\leq k^2, 1\leq k^2$

$$3k^2 + 3k + 1 \leq 3(k^2) + 3(k^2) + 1(k^2) = 7k^2 \leq k^3 \quad\text{when}??$$

edited Jul 07 '13 at 17:31

answered Jul 07 '13 at 17:23

amWhy

209,954

That's where I am stuck. I had myself thought to use induction for this...but am not able to figure out how to prove it. I think Daniel's hint will be useful - though still don't know how. – Parth Thakkar Jul 07 '13 at 17:25
Ok, got it! Thanks – Parth Thakkar Jul 07 '13 at 17:31
You're welcome!! – amWhy Jul 07 '13 at 17:32
@amWhy: Nice feedback +1 – Amzoti Jul 08 '13 at 02:40

score 3 · Answer 3 · answered Jul 07 '13 at 17:37

3

Here's another way. Suppose $k>9$. Then: $$ \begin{align*} k^3&=(k)k^2\\ &>9k^2\\ &=3k^2+3k^2+3k^2\\ &=3k^2+3(k)k+3(k)^2\\ &>3k^2+3(9)k+3(9)^2\\ &=3k^2+27k+243\\ &>3k^2+3k+1\\ \end{align*}$$

answered Jul 07 '13 at 17:37

Adriano

41,576

Awesome. This one is nice too! Although I think it's quite similar to what Daniel and amWhy suggested – Parth Thakkar Jul 07 '13 at 17:38

score 2 · Answer 4 · answered Jul 07 '13 at 17:23

2

If $P(n): 2^n>n^3$

$n=10, P(10): 2^{10}=1024$ and $10^3=1000$

Let $P(n)$ is true for $n=m\implies 2^m>m^3$

Now, $P(m+1): 2^{m+1}=2\cdot 2^m>2m^3$ which we need to be $>(m+1)^3$

$\implies 2>\left(1+\frac1m\right)^3 $

If $m=2, \left(1+\frac1m\right)^3=\frac{27}8>2$

If $m=3, \left(1+\frac1m\right)^3=\frac{64}{27}>2$

If $m=4, \left(1+\frac1m\right)^3=\frac{125}{64}<2$

$\implies 2>\left(1+\frac1m\right)^3$ for $m=4$

and $\left(1+\frac1m\right)^3>\left(1+\frac1{m+1}\right)^3\implies 2>\left(1+\frac1m\right)^3$ for $m\ge 4$

answered Jul 07 '13 at 17:23

lab bhattacharjee

274,582

I don't understand what you wrote after the $P(m+1)$ step - where did you get that implication from? – Parth Thakkar Jul 07 '13 at 17:27
@ParthThakkar, $P(m+1): 2^{m+1}>(m+1)^3$ and from $P(m),2^{m+1}=2(2^m)>2(m^3)$. SO it's sufficient to prove $2m^3>(m+1)^3\iff 2>(1+\frac1m)^3$ – lab bhattacharjee Jul 07 '13 at 17:28
Aah... didn't see that. I don't understand how I can get so silly!! – Parth Thakkar Jul 07 '13 at 17:30
@ParthThakkar, please get back to me if you have any doubt, query – lab bhattacharjee Jul 07 '13 at 17:31
No, I understood this. Thanks! Just because I didn't notice the $P(n)$ step, I got confused, else this is quite straight forward. – Parth Thakkar Jul 07 '13 at 17:35
Wait, I remember that on dividing inequalities, we reach some absurdity. I know that multiplication of inequalities is always valid (i.e. when the sign permits). So, let me do the transition from $P(m)$ to $P(m+1)$ by multiplication. $P(m) \implies 1/2^m < 1/m^3$. But since the sign reverses, we can't multiply the $P(m)$ and $P(m+1)$ equations. So, how is it working? – Parth Thakkar Jul 07 '13 at 17:43
For example of getting absurdity on diving inequalities, consider $ 2 < 3$ and $4 < 6$. – Parth Thakkar Jul 07 '13 at 17:44
@ParthThakkar, $P(m): 2^m>m^3$ is a statement. $2^{m+1}=2\cdot 2^m>2m^3$ if we assume $P(m)$ is true. Now, if we can prove $2m^3>(m+1)^3,$ we are done – lab bhattacharjee Jul 07 '13 at 17:45
let us continue this discussion in chat – lab bhattacharjee Jul 07 '13 at 17:50

score 2 · Answer 5 · answered Jul 07 '13 at 17:27

2

Hint: You can use that if $k\ge10$, then $k^3\ge 10k^2=3k^2+7k^2$.

(Another approach would be to use that $\frac{(k+1)^3}{k^3}=(1+\frac{1}{k})^3\le (11/10)^3<2$ for $k\ge10$.

answered Jul 07 '13 at 17:27

user84413

27,211

The first approach works nicely! – Parth Thakkar Jul 07 '13 at 17:29

score 2 · Answer 6 · answered Jul 07 '13 at 20:58

Your problem, $2^n > n^3$, is equivalent to $n < 2^{n/3}$.

Suppose $n < 2^{n/3}$. Then $2^{(n+1)/3} =2^{1/3}2^{n/3} > n 2^{1/3} $ and $n 2^{1/3} > n+1 \iff n (2^{1/3}-1) > 1 \iff n > \frac1{2^{1/3}-1} \iff n > 3.847... $.

So, if $n \ge 4$ and $n^3 < 2^n$, then $(n+1)^3 < 2^{n+1}$.

Since $1000 = 10^3 < 2^{10} = 1024$, $n^2 < 2^n$ for $n \ge 10$.

Proving by induction: $2^n > n^3 $ for any natural number $n > 9$

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