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Math Help - Converges or diverges - find its limit

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    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
    Last edited by icelated; April 21st 2012 at 06:44 PM.
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    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.
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    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?
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    Re: Converges or diverges - find its limit

    Quote Originally Posted by icelated View Post
    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*}
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