Series of 2^(1/n)1 with limit comparison test, calculus 2 tutorial  Summary and Q&A
TL;DR
The content explains the use of limit comparison test to determine the divergence of the given series.
Questions & Answers
Q: Why is the test for divergence inconclusive in this case?
The test for divergence involves determining the limit as n approaches infinity. In this case, the limit is 0, which does not provide enough information to conclude whether the series converges or diverges.
Q: How is the limit comparison test applied in this analysis?
The limit comparison test involves comparing the given series with a known divergent series, such as the harmonic series. By taking the limit as n approaches infinity, the comparison is made to determine if the series also diverges.
Q: How is L'Hopital's rule used in the analysis?
L'Hopital's rule is used to evaluate the limit of a 0/0 situation. In this case, the rule is applied to determine the limit of 2 to the power of 1/n as n approaches infinity, resulting in a value of ln(2). This value is then used in the limit comparison test.
Q: What is the final conclusion reached in the analysis?
The series represented as Sigma n goes from 1 to infinity of the nth root of 2 minus 1 also diverges, determined by using the limit comparison test and comparing it with the harmonic series.
Summary & Key Takeaways

The content introduces the concept of representing nth root as a power in a series.

The test for divergence is attempted, but it fails due to resulting in 0.

The limit comparison test is used to compare the series with the harmonic series, concluding that it also diverges.