# Math Help - Converges or diverges - find its limit

1. ## Converges or diverges - find its limit

Determine whether the sequence converges or diverges. If it converges, find its limit.
n = 1, 2, 3

$\frac{ 2n - 1} {3n^2 + 1}$

Would i begin by using the quotient rule?
since, by looking at it i might have to prove how i know?

Or could i start by finding the sum?
Or, better just take its limit?

$\frac{ 2} {6n} = 0$

im just not sure what to do

2. ## Re: Converges or diverges - find its limit

Is it a sequence as in $\{a_{n}\}_{n=1}^{\infty} = \frac{2n-1}{3n^{2}+1}$ or a series as in $\sum\limits_{n=1}^{\infty}\frac{2n-1}{3n^{2}+1}$?

In the former, the solution is obvious. In the ladder, the comparison/integral test applies, and it is again an easy solution.

3. ## Re: Converges or diverges - find its limit

It doesn't specify if its a sequence or series. I gave you the whole problem. Would i compare it to 1/n?

4. ## Re: Converges or diverges - find its limit

Originally Posted by icelated
Determine whether the sequence converges or diverges. If it converges, find its limit.
n = 1, 2, 3

$\frac{ 2n - 1} {3n^2 + 1}$

Would i begin by using the quotient rule?
since, by looking at it i might have to prove how i know?

Or could i start by finding the sum?
Or, better just take its limit?

$\frac{ 2} {6n} = 0$

im just not sure what to do
Since the question says sequence, we'll assume that it is a sequence. So we need to evaluate

\displaystyle \begin{align*} \lim_{n \to \infty}\frac{2n - 1}{3n^2 + 1} &= \lim_{n \to \infty}\frac{\frac{1}{n^2}\left(2n - 1\right)}{\frac{1}{n^2}\left(3n^2 + 1\right)} \\ &= \lim_{n \to \infty}\frac{\frac{2}{n} - \frac{1}{n^2}}{3 + \frac{1}{n^2}}\end{align*}