💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # Test the series for convergence or divergence.$\displaystyle \sum_{n = 1}^{\infty} (\sqrt[n]{2} - 1)^n$

## converges.

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##### Top Calculus 2 / BC Educators ##### Heather Z.

Oregon State University  ##### Samuel H.

University of Nottingham Lectures

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### Video Transcript

okay for this problem will let these guys b r a in terms and we'LL use the the root test. So for the root tests we look at Ltd's in goes to infinity of the in through of the absolute value of a m. Okay, so that's going to be lemma as n goes to infinity of and threw it of these guys. So the in through of this is just gonna get rid of the end as the exponents hair. So we'll just get the and through of two minus one. So remember that and threw two is just like two to the power of one over in. If you just rewrite it as n goes to infinity one over and goes to zero two, two zero is one. So this will be one minus one, which is zero. And the important part here is that this is something that's less than one So zero, certainly less than one. So we get convergence #### Topics

Sequences

Series

##### Top Calculus 2 / BC Educators ##### Heather Z.

Oregon State University  ##### Samuel H.

University of Nottingham Lectures

Join Bootcamp