# Series question

1. Mar 16, 2008

### daniel_i_l

1. The problem statement, all variables and given/known data
find $$lim_{n \rightarrow \infty} n^n q^{2^n}$$ where |q|<1

2. Relevant equations

3. The attempt at a solution

I know that $$lim_{n \rightarrow \infty} n^{\alpha} q^{n} = 0$$ So it looks as if the limit in question should also be equal to 0. But how can I prove this?
Thanks

Last edited: Mar 16, 2008
2. Mar 17, 2008

### dynamicsolo

I just want to check, but there isn't supposed to be a summation in there, is there? If not, then you mean you're looking for the limit of a sequence, no?

What about looking at the limit of the ratio of a_n+1 / a_n ? The factors that don't go to one as n goes to infinity are (n+1) · q^(2^n). With |q|<1, the second term goes to zero much faster than n+1 grows. The ratio of consecutive terms goes to zero, so the sequence should go to zero. That sound OK?

3. Mar 17, 2008

### Gib Z

Theres always the 'common sense' response: Since it asks to find $$\lim_{n\to \infty} n^n q^{(2^n)}$$ where |q| < 1, the limit must have the same value for all |q|<1. When q=0, the sequence is equal to 0 for all terms =]

EDIT: Alternative reasoning: $$\lim_{n\to \infty} n^n q^{(2^n)} = \lim_{n\to \infty} e^{\ln \left( n^n q^{(2^n) \right)} = \lim_{n\to \infty} e^{( n \ln n + 2^n \ln q)}$$. Since |q|< 1, ln q is some negative constant. We know the natural log grows slower than any power of x, so the first term in the log can not be larger than say, n^2. 2^n is of a higher order, so exponent part goes to negative infinity, and e raised to that is 0. Reword my logic in more rigorous terms please, this is shameful :(

Last edited: Mar 17, 2008