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anonymous
 5 years ago
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.
anonymous
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=3}^{\infty} (Sin ^{2}(e ^{n}))/n^{4/3}\] 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well the denominator shows a pseries...p = 4/3 > 1 and I'm not sure about the numerator... it seems to be going to infiniti...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because the numerator of this function is either 1 or less for any term, where for a P eries the numerator is always 1, this series will come to a point.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So since the numerator is either 1 or less... it converges because p is > 1? Thanks for your help I really appreciate it =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No problem! I just got done with this section myself. Glad to be of help.
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