Polynomials and exponentials: showing that $n^k \le c\cdot a^n$

Question

If $k$ is any real and $a>1$, prove that there exists a $c>0$ such that for any integer $n\ge 1,$ $$ n^k \le c\cdot a^n $$ To forestall any complaints about the imperative nature of this question, let me give a bit of context. I'm not simply interested in finding a proof; I can give the proof since I can find a suitable value of $c$, given $k$ and $a$. The real question I'm asking is,

How can I provide a proof that students in a 100-level Discrete Mathematics course will understand?

In other words, I expect my students to know pre-calc algebra, but nothing more: no limits, no calculus, so no max/min results, (little or) no understanding of power series, no useful identities like Bernouilli's, and so on. Basically, I can only rely on what they bring from their U.S. high school algebra course.

In the past, I've weaseled out by saying that the proof of the very handy result that $n^k=O(a^n)$ is beyond the scope of the course, but when I sat down to update my lecture notes this summer I wondered whether there was some proof that I'd simply missed. Can anyone come up with one?

Answer 1

I have had some success presenting at high-school level telescopic inductive proofs of related results. In particular, I think it is possible to massage the proof below to suitably simple form.

It is simple to prove powers grow faster than polynomials, i.e. $\rm\ c > 1\ \Rightarrow\ c^n > p(n)\ $ for $\rm\ n > n_0.\:$ For such it suffices to show eventually $\rm\ f(n) = c^n/p(n) > 1\:,\: $ or eventually $\rm\ f(n+1)/f(n) > 1 \ $ since, by multiplicative telescopy, $\rm\:f(n)\:$ is a product of these adjacent term ratios, viz.

$$\rm f(0)\ \prod_{k\:=\:0}^{n-1}\, \frac{f(k+1)}{f(k)}\ = \ \ \color{#C00}{\rlap{--}f(0)}\frac{\color{green}{\rlap{--}f(1)}}{\color{#C00}{\rlap{--}f(0)}}\frac{\color{royalblue}{\rlap{--}f(2)}}{\color{green}{\rlap{--}f(1)}}\frac{\phantom{\rlap{--}f(3)}}{\color{royalblue}{\rlap{--}f(2)}}\, \cdots\, \frac{\color{brown}{\rlap{--}f(n-1)}}{\phantom{\rlap{--}f(n-2)}}\frac{f(n)}{\color{brown}{\rlap{----}f(n-1)}}\ =\ \ f(n) $$

If you view the above cancellation as dynamically collapsing together the cancelling adjacent terms, then it is analogous to collapsing the tubes of handheld telescope into a single tube.

But $\rm\ f(k+1)/f(k)\ =\ c\ p(k)/p(k+1)\ \to\ c > 1\ $ as $\rm\ k\to \infty\ $ since for equal degree polynomials $\rm p(x),q(x)\:$ we have $\rm\ lim_{x\to\infty}\: p(x)/q(x) = p_0/q_0 = $ quotient of their leading coefficients, viz.

$$\rm \frac{p_0\ x^d + p_1\ x^{d-1}+\:\cdots\:+p_d}{q_0\ x^d + q_1\ x^{d-1}+\:\cdots\:+q_d}\ =\ \frac{p_0\ + p_1/x +\:\cdots\:+p_d/x^d}{q_0\ + q_1/x +\:\cdots\:+q_d/x^d}\ \to\ \frac{p_0}{q_0}\ \ \ as\ \ \ x\to\infty$$

As I have mentioned here before in many posts, by means of cancelling complicated expressions, telescopy often reduces induction problems to trivialities (here the fact that a product of terms eventually $> 1$ is itself eventually $> 1$). Difficult problems involving hyperrational functions (i.e. $\rm\ f(n+1)/f(n) = $ rational function of $\rm\:n,\:$ such as powers and exponentials) are, after application of telescopy, greatly simplified to trivial problems about rational functions - functions so simple that questions about such can be decided mechanically by algorithms.