# Show $b^n > n$ from first principles [duplicate]

I have to show that for any $$b >1$$, we have $$b^n > n$$ for all $$n$$ sufficiently large, using only very basic analysis (no calculus). My attempt is as follows.

We know that $$b^{n+1} - b^n = b^n(b-1)$$. For $$n$$ sufficiently large, say $$n \geq N = \left\lceil \frac{\ln(2/(b-1))}{\ln b} \right\rceil + 1,$$ we have $$b^{n+1} - b^n > 2.$$

Now let $$\Delta = N - b^N$$. Then for any $$j\geq 1$$, we have $$b^{N+j} = (b^{N+j} - b^{N+j-1}) + \ldots + (b^{N+1} - b^N) + b^N > 2j + b^N = 2j + N - \Delta = N+j + (j - \Delta).$$ Thus we have $$b^n > n$$ for any $$n \geq N + |\Delta|$$.

This works, but it seems messy. Is there is better way? I know induction is usual for this type of problem, but establishing the base case for generic $$b$$ seems difficult.

## marked as duplicate by Martin R, ancientmathematician, mrtaurho, Theo Bendit, GNUSupporter 8964民主女神 地下教會Feb 14 at 13:08

• – Martin R Feb 13 at 20:46

If $$b > 1$$ you can write $$b = 1 + x$$ with $$x > 0$$ and by Bernoulli's inequality $$b^n = (1+x)^n \ge \frac{n(n-1)}{2}x^2.$$ Thus $$\frac{b^n}{n} \ge \frac{(n-1)x^2}{2} \ge 1$$ for all $$n > \dfrac{2}{x^2} + 1.$$

• That's math.stackexchange.com/a/1738428/42969 from the possible duplicate target :) – Martin R Feb 13 at 20:47
• Nuts. I thought I was being clever too :( – Umberto P. Feb 13 at 20:50
• I thought Bernoulli's inequality says $(1+x)^n \ge 1+nx$ – J. W. Tanner Feb 13 at 21:02
• Another version of Bernoulli's inequality states $(1+x)^n \ge 1 + nx + \frac{n(n-1)}{2}x^2$. – Umberto P. Feb 13 at 21:05
• And you could leave out $1+nx$ because $1+nx > 0$ ? – J. W. Tanner Feb 13 at 21:25

(Umberto P.'s argument from first principles is surely the best way to go, but here's another argument, just for the sake of variety.)

Consider the sequence $$a_n = \frac{b^n}{n}$$. We have: $$\frac{a_{n+1}}{a_n} = \frac{b}{1 + \frac{1}{n}} \geqslant \frac{2b}{b+1} > 1 \text{ for } n \geqslant \frac{2}{b-1}.$$ If the eventually increasing sequence $$(a_n)$$ is bounded above, it tends to a limit, but this implies $$1 = \lim_{n\to\infty}\frac{a_{n+1}}{a_n}\geqslant \frac{2b}{b+1} > 1,$$ a contradiction. Therefore, $$(a_n)$$ is not bounded above. In particular, there exists $$N \geqslant \frac{2}{b-1}$$ such that $$a_N > 1$$. Then we have $$a_n > 1$$ for all $$n \geqslant N$$.

We can proceed by induction. Suppose $$b^x>x$$ is true for all $$x= n$$. We want to show the claim for $$x=n+1$$. We just need $$b^n(b-1)=b^{n+1}-b^n>1,$$ Because by inductive hypothesis we have $$b^n>n$$; then we can sum the inequalities. But $$b^n(b-1)>1$$ is the same as $$b^n>1/(b-1)$$, which is clearly true when $$n$$ is sufficiently large, since $$b^n$$ grows without bound but $$b-1$$ is constant.