# Prove lim na^n = 0 when 0 < a < 1

1. Aug 13, 2010

### e(ho0n3

The problem statement, all variables and given/known data
Prove lim na^n = 0 when 0 < a < 1.

The attempt at a solution
Without danger, we change from the discrete n to the continuous x so that now we have to prove that lim xa^x = 0. Let e > 0. We have to find an N such that xa^x < e for all x > N. Now if xa^x < e is the same as 1 < eb^x/x, where b = 1/a. Using L'Hospital's rule, we have that lim eb^x/x = lim e(ln b)b^x = oo, so there is an N such that 1 < eb^x/x. QED

Is this the simplest way to prove this limit. For some odd reason, I feel that there is a simpler solution. Any tips?

2. Aug 13, 2010

### diegocas

Apply L'Hopital's rule in the limit $$\lim x.a^x$$ after writing $$x.a^x = \frac{x}{a^{-x}}$$.

3. Aug 13, 2010

### Hurkyl

Staff Emeritus
My first inclination is not to use calculus at all!

Let a < u < 1. Prove that, eventually, (n+1)/n < u/a. Then, an easy inductive proof shows that 0 < n a^n < C u^n for some constant C.

I suppose technically I've used some amount of calculus in that I invoke the Archimedean property, the squeeze theorem, and knowledge that u^n converges to 0.

Come to think of it, the inductive step isn't really going to be conceptually much different from taking a derivative or a logarithmic derivative. (Though the analogy might seem opaque if you're just learning this stuff)

4. Aug 13, 2010

### Dick

I would do it by showing log(n*a^n)/n goes to log(a) using l'Hopital. Hence log(n*a^n) approaches -infinity.

5. Aug 14, 2010

### e(ho0n3

Thank you for the suggestions. I'm a little rusty on this stuff.