# Thread: finding the limit of this function...

1. ## finding the limit of this function...

After staring at this problem forever, I'm utterly at a loss as to how to approach this problem.

the limit as n goes to infinity of (n^n)/[n!(2^n)]

It has to do with whether the numerator or the denominator goes to infinity "faster" but I don't know how to start this problem off...

2. Originally Posted by choe83188
After staring at this problem forever, I'm utterly at a loss as to how to approach this problem.

the limit as n goes to infinity of (n^n)/[n!(2^n)]

It has to do with whether the numerator or the denominator goes to infinity "faster" but I don't know how to start this problem off...

use the ratio test to prove that the series $\sum_{n=1}^{\infty} \frac{2^n n!}{n^n}$ converges and so $\lim_{n\to\infty} \frac{2^n n!}{n^n} = 0.$ thus the limit of your sequence is $\infty.$

3. Originally Posted by NonCommAlg
use the ratio test to prove that the series $\sum_{n=1}^{\infty} \frac{2^n n!}{n^n}$ converges and so $\lim_{n\to\infty} \frac{2^n n!}{n^n} = 0.$ thus the limit of your sequence is $\infty.$

Thanks much, hadn't thought of that. Will be doing that now.

Thanks again!