💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# Determine whether the sequence converges or diverges. If it converges, find the limit.$a_n = \sqrt [n]{2^{1 + 3n}}$

## The sequence converges to 8

Sequences

Series

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

So I think the easiest way to approach this problem is by rewriting this A in term. So this in that we see happening over there, That just means that all this stuff is being taken to the one over in power. So the one over n And then as long as you know how exponents work, then you know that this one over in is going to distribute to all of this stuff. So this turns into one times one over in plus three in times one over in, which is just three. So now if we do limit as n goes to infinity of a n mhm, we get limit as n goes to infinity of two to the one over N Plus three and to to the X is a continuous function, which means that you're allowed to pull the limit inside of the function. So this is the same thing as two to the power of limit as n goes to infinity of one over n plus three As n goes to infinity, one over N goes to zero and we get two to the three, which is eight. So our sequence converges and it converges to eight. Yeah,

Sequences

Series

Lectures

Join Bootcamp