Comparing Series Divergence: \(\sum\frac{n^2-\arctan(n)}{n^3+\sin(n)}\)

[Bachelier]

\sum (n^2-arctan(n)) / (n^3 + sin(n)) n=0 to ∞

I know this series diverges, but how would you use the comparison test to compare it to (n^2 / n^3 meaning the harmonic series 1/n)

Thank you very much

 

[HallsofIvy]

|sin(n)|\le 1| so the denominator can be replaced by n^3+ 1> n. 0\le arctan(n)\le 2\pi so the numerator can be replaced by n^2- 2\pi< n^2.

 

[Bachelier]

|sin(n)|\le 1| so the denominator can be replaced by n^3+ 1> n. 0\le arctan(n)\le 2\pi so the numerator can be replaced by n^2- 2\pi< n^2.

Thanks for the reply

I thought - \pi/2\le arctan(n)\le \pi/2 for one

and two, based on your reasoning, our series will end up being smaller than n^3/n^2 = n which in turn diverges, therefore this test is inconclusive.

And the denominator n^3 + sin(n) will make the whole series bigger that a series with a denominator of n^3+ 1 but having a numerator that is smaller than the numerator of the second series with the denominator of n^3+ 1, will make it hard to decide which series is bigger.

thank you

 

[Dick]

Call the nth term of the series an. Then you know limit n*an as n->infinity is 1, right? Compare the series with 1/(2n) for sufficiently large n.