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Series, find Divergence or Convergence

  1. May 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the Divergence or Convergence of the series

    [itex]\sum[/itex][itex]^{∞}_{n=1}[/itex][itex]\frac{2n^2+3n}{\sqrt{5+n^5}}[/itex]

    2. Relevant equations

    Ratio Test, Comparison Test, Limit Comparison Test, Integral test etc.

    3. The attempt at a solution

    This question was on my final exam and the only question of which I couldn't actually figure out. I tried the ratio test multiple times and ended getting L=1 which is no info multiple times. I figured the comparison test was the right approach but had no idea what series to compare it to.
     
  2. jcsd
  3. May 18, 2012 #2

    sharks

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    You could compare it to: [tex]\frac{2n^2}{\sqrt{n^5}}[/tex]Or you could try factorizing it by taking out [itex]n^2[/itex] from the numerator and denominator.
     
    Last edited: May 18, 2012
  4. May 18, 2012 #3
    Can you expand on your suggestions? I am utterly clueless with respect to this problem.
     
  5. May 18, 2012 #4

    sharks

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    Taking out [itex]n^2[/itex] from the numerator and denominator:[tex]\sum^{\infty}_{n=1}\frac{2+\frac{3}{n}}{\sqrt{5/n^4 + n}}[/tex]Well, the nth-term test is not useful here, as the limit is 0.
     
  6. May 18, 2012 #5
    Ye I was about to say that lol.
     
  7. May 18, 2012 #6

    sharks

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    You should always try with the simpler tests first. :smile:

    Using the comparison test: [tex]u_n=\frac{2n^2+3n}{\sqrt{5+n^5}}[/tex][tex]v_n=\frac{2n^2}{\sqrt{n^5}}=\frac{2}{n^{1/2}}[/tex]Now, use the p-series test and the series ##v_n## diverges.
    Since [itex]v_n \le u_n[/itex] and [itex]\sum^{\infty}_{n=1}v_n[/itex] diverges, therefore the original series diverges.
     
  8. May 18, 2012 #7
    Thanks.
     
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