Proving $\forall n \geq10, n^3 < 2^n$

Question

Im asked to prove the following:

$$\forall n \geq10, n^3 < 2^n$$

I'm 100% sure that the easiest way to prove this is via induction but I'm not being able to do so. How can I prove this?

Answer 1

Proof without induction : The function $$f : x \mapsto \frac{2^x}{x^3}$$

is differentiable over $[10, +\infty)$, and $$f'(x)=\frac{2^x\left[\ln(2)x - 3\right]}{x^4} > 0$$

for all $x \geq 10$. Therefore $f$ is strictly increasing over $[10, +\infty)$, so for $n \geq 10$, you have $$f(n) \geq f(10) = \frac{2^{10}}{10^3} = \frac{1024}{1000} > 1$$

i.e. for all $n \geq 10$, you have $$2^n > n^3$$

Answer 2

To show that $(n+1)^3 < 2n^3$, you can also use the fact that for $n \ge 10$: $$\dfrac {(n+1)^3}{n^3} = \left(1+\dfrac1n\right)^3 \le \left(1+\dfrac1{10}\right)^3 < 2$$

Answer 3

For $n=10$, it is obvious that $2^{10}>10^3$

Assume that for $k\geq 10, \space 2^k>k^3$

Prove for $n=k+1$

$$2^{k+1}>(k+1)^3$$

So

$$2^{k+1}=2\times2^k>2k^3=k^3+k^3>k^3+3k^2+3k+1=(k+1)^3$$

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

$$k\geq10>3+3+1$$

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

Answer 4

Let $P(n):\ n^3<2^n$.

For the base case $n=10$, we have $P(10):\ 1000<1024$ which is true.

Let us assume that $P(k)$ is true for some $\mathbb{N}\ni k\geq 10$. Then, $$k^3<2^k\quad (*)$$ Now, we shall prove that $P(k+1)$ is also true. Note that $2^{k+1}=2\cdot 2^k$. Using $(*)$, we get $$2^{k+1}>k^3+k^3\quad (**)$$ The key here is to note that for $n\geq 10$, $n^3\geq 1000$ grows much faster than $3k²+3k+1\geq 331$. If required, this can be proved easily.

Now, using the inequality above, $(**)$ becomes $$2^{k+1}>k^3+3k^2+3k+1$$ $$\implies 2^{k+1}>(k+1)^3$$ Thus, we have proved that $P(k+1)$ is true whenever $P(k)$ is true.

Hence, by the principle of mathematical induction, we conclude that $P(n)$ is true for all $\mathbb{N}\ni k\geq 10$.

Hope this helps :)

Answer 5

We have that

$$n^3 < 2^n \iff \sqrt[n] n < \sqrt[3] 2$$

and we can check the cases $n=10, 11$ by inspection and then using that for $n\ge 12$

$$\sqrt[n] n =e^{\frac{\log n}{n}} \le1+\frac{\log n}{n}+\left(\frac{\log n}{n}\right)^2\le 1+\frac{\log 12}{12}+\left(\frac{\log 12}{12}\right)^2 <\sqrt[3] 2$$