What test do I use to see if the series is divergent or convergent?

Question

$$\sum^{\infty}_{n = 1}{3n + 1 \over n^{2} - n + 2}$$

What test do I use to find out if the limit is convergent or divergent?

Answer 1

Expounding on Mehta's answer..

Using Limit Comparison Test

For $\sum a_n$ and $\sum b_n$, series with positive terms, if

$$\lim_{n \to \infty} \frac{a_n}{b_n} \ne 0$$

where if $\sum b_n$ converges/diverges $\implies$ $\sum a_n$ converges/diverges respectively.

$$a_n = \frac{n}{n^2} \cdot \frac{3 + \frac{1}{n}}{1 - \frac{1}{n} + \frac{2}{n^2}} \approx \frac{1}{n} \cdot \frac{3}{1}$$

For $n$ very very large $\frac{1}{n}$, $\frac{2}{n^2} \approx 0$

Take $b_n = \frac{1}{n}$

$$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n\to\infty}\frac{\frac{3n+1}{n^2-n+2}}{\frac{1}{n}} = \lim_{n\to\infty} \frac{3n+1}{n^2-n+2} \cdot n = \lim_{n\to\infty} \frac{3n^2 + n}{n^2-n+2}$$

Divide through by the highest power of $n$..

$$\lim_{n\to\infty} \frac{\frac{3n^2}{n^2} - \frac{n}{n^2}}{\frac{n^2}{n^2} - \frac{n}{n^2} + \frac{2}{n^2}} = \lim_{n\to\infty} \frac{3 -0}{1 -0+0} = \frac{3}{1} \ne 0$$

$\frac{n}{n^2}$, $\frac{2}{n^2}$ $\to 0$ as $n \to \infty$

$\sum \frac{1}{n}$ is divergent because it is a p-series with $p =1$ (if $\sum \frac{1}{x^p}$, and $p \le 1$then the series diverges)

Since $$\lim_{n \to \infty} \frac{a_n}{b_n} = 3 \ne 0$$

And we know $\sum \frac{1}{n}$ diverges, $$\sum_{n=1}^\infty \frac{3n+1}{n^2-n+2}$$ diverges.

Answer 2

Try the limit comparison test. $\dfrac{3n+1}{n^2 - n + 2}$, for large values of $n$, looks sort of like $\dfrac{3n}{n^2} = \dfrac{3}{n}$. So you can try the limit comparison test with $\dfrac{3}{n}$.