Prove the series converges or diverges $\sum_{r=1}^{\infty}\frac{2r-1}{(3r^{3}-2)}$ . $$\sum_{r=1}^{\infty}\frac{2r-1}{(3r^{3}-2)}.$$
I am trying to use the comparison test and get stuck at $2j-1/(3j{^3} -2)>2j-2/(3j{^3})$ but I feel like I'm making things worse.
 A: Use the comparison test (limit form).
Take $\sum v_r=\sum\frac{1}{r^2} $, $\sum u_r=\sum\frac{2r-1}{3r^{3}-2}$
Now , $\lim_{r \to \infty} \frac{u_r}{v_r}$
$= \lim_{r \to \infty} \frac{2r^{3}-r^{2}}{3r^{3}-2} $
$= \frac{2}{3} $.
So,  $\sum_{r=1}^{\infty} \frac{2r-1}{3r^{3}-2}$ converge , as $\sum_{r=1}^{\infty} \frac{1}{r^2} $ converge.
A: If you want to use the comparison test, then you have to ensure that your new sequence is greater than the current sequence $\displaystyle \sum_{r=1}^{\infty} \frac{2r-1}{3r^3-1}$.
We know that $2r>2r-1$ and $3r^3-3<3r^3-2$, therefore $\displaystyle \frac{2r-1}{3r^3-2}<\frac{2r}{3(r^3-1)}=\frac{.67}{r^2-1/r}< \frac{1}{r^2}$ because both $\dfrac{1}{r^2}, \dfrac{1}{r^2-1/r}$ converge to zero.
And we know that the $\displaystyle \sum \frac{1}{r^2}$ converges, therefore by the comparison which states that if $a_n<b_n$ for all n then $\sum a_n<\sum b_n$ which means in $\sum b_n$ converges then $\sum a_n$ converges.
A: You have
$$\frac{2r-1}{3r^3-2} \sim \frac{2}{3r^2}$$
which is the general term of a convergent series, therefore your series converges.
