Convergence of Sequence xn=nabn and Lim of P(n)/(a^n) = 0

[RKermanshahi]

1. Consider the sequence x_n=n^abⁿ, where a is a natural number and b is a real number with 0<b<1. Show that the sequence converges to zero. Conclude from here that lim P(n)/cⁿ=0, where P is a polynomial function and c>1.



2. I am not sure how to show that the sequence converges to zero.


3. We know that bⁿ certainly converges to zero since 0<b<1. I have tried to show that since that part of the sequence converges to zero, the entire sequence converges to zero; however, I believe I need that at least the rest of the sequence is bounded, which it is not. I have also tried using the standard epsilon-definition of the convergence of a sequence, but that has proved to be messy, with ln's and e's. My guess is it's something simple that I'm not seeing...

 

[Dick]

I would try to work with log(x_n)=log(n^a*b^n). Can you show the log goes to negative infinity as n->infinity? You can do this if you can show log(x_n)/n goes to log(b) as n->infinity.

 

[RKermanshahi]

So taking the log's of both sides and applying rules of log's we get:

log(x_n)/n= (alogn)/n + logb. Is there a way to show (a logn)/n goes to zero as n goes to infinity without using L'Hospital's rule?

 

[Dick]

So taking the log's of both sides and applying rules of log's we get:

log(x_n)/n= (alogn)/n + logb. Is there a way to show (a logn)/n goes to zero as n goes to infinity without using L'Hospital's rule?

Nothing very neat, I guess. You could show the limit of log(n)/n is the same as the limit of log(e^n)/e^n, since e^n also goes to infinity. So that gives you n/e^n. If b_n=n/e^n then the limit of (b_n+1)/(b_n) is 1/e which is less than 1. So you can show it goes to zero by comparison with (1/e)^n.

 

[RKermanshahi]

Thanks, Dick. I actually solved it using the monotone convergence theorem. I took the quotient x_n/x_n+1 and showed that it was eventually always less than one (and hence eventually decreasing). I then showed it was bounded. So by the monotone convergence theorem it converges. Finally, taking the limit of the recurrence relation, we get that the limit equals 0!