## Calculus 8th Edition

$f(x)=\Sigma_{n=3}^{\infty} \frac{3n-4}{n^{2}-2n}$ $\Sigma_{n=3}^{\infty} \frac{3n-4}{n^{2}-2n}=\Sigma_{n=3}^{\infty} \frac{3n-4}{n(n-2)}\gt \Sigma_{n=3}^{\infty} \frac{3n-4}{n.n}$ 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.