Show $b^n > n$ from first principles

Question

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.

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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 $n \geq N + |\Delta|$.

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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.

Answer 1

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 $n > \dfrac{2}{x^2} + 1.$

Answer 2

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.

Answer 3

(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 $n \geqslant N$.