anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\sum_{n=3}^{\infty} (Sin ^{2}(e ^{n}))/n^{4/3}\] 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
  • anonymous
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...
anonymous
  • anonymous
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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anonymous
  • anonymous
because 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
  • anonymous
So since the numerator is either 1 or less... it converges because p is > 1? Thanks for your help I really appreciate it =)
anonymous
  • anonymous
No problem! I just got done with this section myself. Glad to be of help.

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