## 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.

1. 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.

2. 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...

3. 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.

4. 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.

5. anonymous

So since the numerator is either 1 or less... it converges because p is > 1? Thanks for your help I really appreciate it =)

6. anonymous

No problem! I just got done with this section myself. Glad to be of help.