# How do I determine convergence with a comparison test?

$$\sum_{n=2}^\infty \frac{1}{3n^2-2\sqrt{n}}$$

The instructions are to use a comparison test to determine convergence of the series. I thought to compare it to

$$\sum_{n=2}^\infty \frac{1}{3n^2}$$

which converges by the p-series test. But, that series is smaller than the original series, so it doesn't prove that the original series will converge too. Where do I go from here?

• You already get the idea the series is 'roughly' similar to a $2$-series. So if $\dfrac{1}{3n^2}$ is too small, can you find another constant to make the $2$-series larger than $\displaystyle\sum^{+\infty}_{n=2}\dfrac{1}{3n^2-2\sqrt{n}}$? Commented Mar 30 at 6:10
• What part of the series is the 'constant' and what does it mean to be a '2-series'? Commented Mar 30 at 6:18
• We call $\displaystyle\sum^{+\infty}_{i=1}\dfrac{1}{n^p}$ be a $p$-series. In the case $p=2$, so we call it a $2$-series. For the constant, I actually mean the $\dfrac{1}{3}$ is making the $2$-series too small, so asking you to suggest a new candidate of $2$-series but not with $\dfrac{1}{3}$. Commented Mar 30 at 6:20
• Okay, I understand why we call it a 2-series. But I'm struggling to understand the other part. How does choosing a different constant help, and how do I know what to choose? Commented Mar 30 at 6:31
• For choosing constants, the idea is like, you know you need a constant larger than $\dfrac{1}{3}$, so you can try like $1$, $2$, etc. But how do you prove that? Usually either by induction, or it maybe quite obvious (like the hints below). So the choice is quite depending on you, because let's say $1$ is suitable choice, then every number greater than $1$ is also suitable. Commented Mar 30 at 6:31

Hint: $$3n^2-2\sqrt{n}\ge n^2$$ for $$n\ge 2$$.
$$3n^2-2\sqrt{n}\ge n^2\iff 2n^2\ge 2\sqrt{n}\iff n^4\ge n\iff n\ge1$$. So we have $$\sum^m_{n=2}\dfrac{1}{3n^2-2\sqrt{n}}\le\sum^m_{n=2}\dfrac{1}{n^2}$$for each $$m\ge2$$. By $$\cdots$$