Originally Posted by

**pirateboy** Here' we're asked to determine whether the sequence converges or diverges, and if it converges we are asked to find the limit. We are given the sequence

$\displaystyle a_n = \sqrt[n]{2^{1+3n}}$

So I can see by putting this into my calculator that this is going to converge to 8. My problem is showing this mathematically. Here's where I'm stuck.

$\displaystyle a_n = \sqrt[n]{2^{1+3n}} = \left(2^{1+3n}\right)^{1/n}$

Easy enough, now the tough part....

$\displaystyle \displaystyle \lim_{n \to \infty}\left(2^{1+3n}\right)^{1/n}$

Where do I even begin on this? Do I need to use a natural log or something? Wood that be the thing to do? (terrible pun, but I had to do it.)

I obviously can't say that my exponent is going to 0 as n goes to infinity, because that's just not true. If it were the answer would just be 1 and that'd be it, but that's obviously not the case.

If anyone could help point me in the right direction, that'd just be super.

Thanks!