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.