I'm stuck on this series question. Trying to figure out what to compare it to: series from (n= 3 to infiniti) of (sin(e^n))^2/n^(4/3) Am I doing wrong by comparing it to e^n/n^(4/3)??? Thanks.

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\[\sum_{n=3}^{\infty} (Sin ^{2}(e ^{n}))/n^{4/3}\] e^n / n^(4/3) would seem to be a good choice. Explore the limit of sin^2(e^n) as N goes to infinity.

Well the denominator shows a p-series...p = 4/3 > 1 and I'm not sure about the numerator... it seems to be going to infiniti...

The Sum of the numerator will go to infinity, yes, however the limit of the sine function is from -1 to 1. When you square it you limit the range of the sine function from 0 to 1. The numerator, therefore, can never be greater than 1. A p series converges if p is greater than 1 and the numerator is 1. So, in this case, the numerator is either 1 or something less.

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