Answer
Divergent
Work Step by Step
$f(x)=\Sigma_{n=3}^{\infty} \frac{3n-4}{n^{2}-2n}$
$\Sigma_{n=3}^{\infty} \frac{3n-4}{n^{2}-2n}=\Sigma_{n=3}^{\infty} \frac{3n-4}{n(n-2)}\gt \Sigma_{n=3}^{\infty} \frac{3n-4}{n.n}$
Note that $\Sigma_{n=3}^{\infty} \frac{3n-4}{n.n}=\Sigma_{n=3}^{\infty} \frac{3}{n}-\frac{4}{n^{2}}$
$=$($p$-series with $ p=1$)$-(p$-series with $p=2\gt 1$)
= Diverging - converging series
$=\infty$
Thus, $\Sigma_{n=3}^{\infty} \frac{3n-4}{n(n-2)}\gt \infty $
Hence, the given series is divergent.
