Convergence test examples

Example

Consider

$$\sum _{n=2}^{\infty }\frac{3n+\sin (n)}{n^3-n^2}$$

We compare with the p-series $\sum \frac{1}{n^2}$.

Since $|\sin (n)|\le 1$,

$$|3n+\sin (n)|\le 3n+1\le 4n(n\ge 1)$$

Also,

$$n^3-n^2=n^3\left(1-\frac{1}{n}\right)$$

For $n\ge 2$,

$$1-\frac{1}{n}\ge 1-\frac{1}{2}=\frac{1}{2}$$

so we can take $c=\frac{1}{2}$ and get

$$n^3-n^2\ge \frac{1}{2}{n}^3(n\ge 2)$$

Hence for $n\ge 2$,

$$\left|\frac{3n+\sin (n)}{n^3-n^2}\right|\le \frac{4n}{\frac{1}{2}{n}^3}=\frac{8}{n^2}$$

Since ${\sum }_{n=2}^{\infty }\frac{8}{n^2}$ converges by the p-series test, the comparison test gives

$$\sum _{n=2}^{\infty }\frac{3n+\sin (n)}{n^3-n^2}$$

converges absolutely.