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Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $

I did these steps:

  1. Basis step
    $$P(10): \ \ 1024 \geq 1000 \ (True)$$
  2. Inductive step $$P(n) \implies P(n+1) \\P(n+1) = 2^{n+1} \geq (n+1)^3$$ so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3 \cdot 2 \\ 2^{n+1} \geq 2n^3 \\ 2^{n+1} \geq (n+1)^3$$ taking advantage of the fact that $(n+1)^3 \geq 2n^3$ and less than $2^{n+1}$.

does this demonstration work?

I know there's a similar question but the solution is different, I want to know if my demonstration is valid as well.

Loris Simonetti
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2 Answers2

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so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3 \cdot 2 \\ 2^{n+1} \geq 2n^3 \\ 2^{n+1} \geq (n+1)^3$$ taking advantage of the fact that $(n+1)^3 \geq 2n^3$ and less than $2^{n+1}$.

does this demonstration work?

No, your analysis (in and of itself) hasn't shown that
$(n+1)^3 \leq 2^{n+1}$, assuming that I am not overlooking anything. This is what you are trying to prove.

The easy way, is:

after noting that $2^{10} > 10^3$, and assuming that $2^n > n^3$, simply note that for $n \geq 10$,

$$\left(\frac{n+1}{n}\right)^3 < 2. \tag1$$

Note that $2^{1/3} > 1.25$, and for $n \geq 10, \frac{n+1}{n} < 1.25$.

This justifies equation (1) above.

Thus:

$$2^{n+1} = 2 \times 2^n > 2 \times n^3 > \left(\frac{n+1}{n}\right)^3 \times n^3 = (n+1)^3.$$

jjagmath
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user2661923
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  • I didn't understand how you found $$ [\frac{n+1}{n}]^3 $$ because 'n' can be simplified with $n^3$? – Loris Simonetti Jan 23 '21 at 12:25
  • @LorisSimonetti I asked myself: when starting with the inequality $$2^n > n^3$$ and attempting to use that to prove that $$2^{(n+1)} > (n+1)^3,$$ what is happening to each side of the inequality? The LHS is being multiplied by $2$ and the RHS is being multiplied by $\left[\frac{n+1}{n}\right]^3.$ – user2661923 Jan 23 '21 at 12:30
  • After applying the induction hypothesis, you want that $2n^3$ to be greater than $(n+1)^3$ – jjagmath Jan 23 '21 at 12:31
  • @jjagmath which is equivalent to what I proved, which is that $$2 > \left[\frac{n+1}{n}\right]^3.$$ – user2661923 Jan 23 '21 at 12:33
  • @user2661923 Ok, got it! thanks. – Loris Simonetti Jan 23 '21 at 12:34
  • @user2661923 Yes, I was answering to Loris, but you answered first – jjagmath Jan 23 '21 at 12:34
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You already proved the base case

Now suppose that $n^3\le 2^n$ is true and

let's prove it for $n+1$

$(n+1)^3=n^3+\left(3n^2+3n+1\right)\tag{1}<\ldots$

for $n\ge 4$ we have $3n^2+3n+1 < n^3$ indeed, adding $n^3$ to both sides we get

$n^3+3 n^2+3 n+1<n^3+n^3\to (n+1)^3<2n^3\to n+1<n\sqrt[3]{2}\to n(\sqrt[3]{2}-1)>1\to n>\frac{1}{\sqrt[3]{2}-1}\approx 3.8$

continue $(1)$ $$\ldots<n^3+n^3\le 2^n+2^n=2^{n+1}$$

Raffaele
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