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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.

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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

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

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

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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.

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