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Calculating a limit of a sequence

  1. Jan 22, 2012 #1
    Hi guys,

    1. The problem statement, all variables and given/known data
    I'm trying to calculate the limit of the following sequence:
    an:= ([itex]\frac{n^{2}+10}{n^{2}-5n}[/itex])[itex]^{(3n^{2}+2)}[/itex]​

    2. The attempt at a solution
    Ok so I got this so far:

    an = [itex]\left(\frac{\left(n^2\left(1+\frac{10}{n^2}\right)\right)}{\left(n^2\left(1-\frac{5}{n}\right)\right)}\right)^{n^{2}(3+\frac{2}{n})}[/itex]​

    [itex]\lim_{n\to \infty }[/itex] an = [itex]\lim_{n\to \infty }[/itex] 1[itex]^{∞}[/itex]​

    So this is what I got till now and I don't know really how to proceed or how can I approach this problem differently to find the correct solution which is ∞.

    Any help, tips are welcome.
  2. jcsd
  3. Jan 22, 2012 #2


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    Without having worked the problem out, my first thought is to find the limit of the log of this.

    If [itex]\displaystyle \lim_{n\to\infty}\ln(a_n)=L\,,\ \text{ then } \lim_{n\to\infty}a_n=e^L[/itex]
  4. Jan 23, 2012 #3
    I don't know if this is allowed with products of limits rule but:

    [tex]3n^2 \ln \left( \frac{n^2}{n-5n}\right) = 3n \ln \left[\left( 1 +\frac{5}{n} \right)^n \right] = 3n \cdot 5 = 15n \rightarrow \infty [/tex]
  5. Jan 23, 2012 #4


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    Binomial theorem.

    [itex]{(\frac{n^{2}+10}{n^{2}-5n})}^{(3n^{2}+2)} = {(1 + \frac{5}{n} + \frac{35}{n(n-5)})}^{(3n^2 + 2)} > {(1 + \frac{5}{n})^{3n^2} = {[{(1+\frac{5}{n})}^n}]}^{3n}[/itex], the limit of which is [itex]e^{15n}[/itex], which is ∞.

    This is a slight (more direct) variation of dirk_mec's idea.

    EDIT: BTW, I messed up the sign, clearly it should be greater than ('>').
    Last edited: Jan 23, 2012
  6. Jan 23, 2012 #5
    Thank you guys for your help and I'm posting my solution to the problem.

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