Series of 2^(1/n)-1 with limit comparison test, calculus 2 tutorial | Summary and Q&A

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June 1, 2016
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blackpenredpen
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Series of 2^(1/n)-1 with limit comparison test, calculus 2 tutorial

TL;DR

The content explains the use of limit comparison test to determine the divergence of the given series.

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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.

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