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

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?

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  • $\begingroup$ 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 $\displaystyle\sum^{+\infty}_{n=2}\dfrac{1}{3n^2-2\sqrt{n}}$? $\endgroup$
    – Angae MT
    Commented Mar 30 at 6:10
  • $\begingroup$ What part of the series is the 'constant' and what does it mean to be a '2-series'? $\endgroup$
    – ur500
    Commented Mar 30 at 6:18
  • $\begingroup$ 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 $\dfrac{1}{3}$. $\endgroup$
    – Angae MT
    Commented Mar 30 at 6:20
  • $\begingroup$ 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? $\endgroup$
    – ur500
    Commented Mar 30 at 6:31
  • $\begingroup$ 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. $\endgroup$
    – Angae MT
    Commented Mar 30 at 6:31

1 Answer 1

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Hint: $3n^2-2\sqrt{n}\ge n^2$ for $n\ge 2$.

In case you do not know, I provide more steps below.

$3n^2-2\sqrt{n}\ge n^2\iff 2n^2\ge 2\sqrt{n}\iff n^4\ge n\iff n\ge1$. 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 $\cdots$

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