- #1

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- Thread starter squenshl
- Start date

- #1

- 479

- 4

- #2

- 1,631

- 4

- #3

- 479

- 4

Thought so, thanks.

For the first one I know a(n) = sin^2(1/n), so could I use b(n) = 1/(n^2).

As for the second I see that the limit as n tends to inf using the divergence test is 1/2 which doesn't equal 0 so therefore this series diverges. Thanks

For the first one I know a(n) = sin^2(1/n), so could I use b(n) = 1/(n^2).

As for the second I see that the limit as n tends to inf using the divergence test is 1/2 which doesn't equal 0 so therefore this series diverges. Thanks

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- #4

- 479

- 4

Any help.

And would I use a ratio test on the series n!/(2n+1).

- #5

- 534

- 1

For |sin n|/n

- #6

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Warren.

- #7

- 1,631

- 4

Well, while it is true that many students, including myself, ask for help in these forums, i would say that for the most part the people who seek for help here get only hints. THat is, it is them(with some exceptions) who do the most part of the job.

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