Determine if series 3/n*sqrt(n) converges or diverges.

  1. 1. The problem statement, all variables and given/known data

    [tex]\sum^{\infty}_{n=1}\frac{3}{n^{1+\frac{1}{n}}}[/tex]

    2. Relevant equations
    Comparison Test
    Limit Comparison Test
    Test of Convergence (Just to show it doesn't immediately diverge)


    3. The attempt at a solution
    I sort of just would like to check to make sure I'm getting a proper [tex]b_{n}[/tex]

    Manipulating the Series:
    [tex]\sum^{\infty}_{n=1}\frac{3}{n^{1+\frac{1}{n}}}[/tex]

    [tex]\sum^{\infty}_{n=1}\frac{3}{n^{\frac{n+1}{n}}}[/tex]

    [tex]\sum^{\infty}_{n=1}\frac{3}{n\sqrt[n]{n}}[/tex]

    Test of Convergence:
    limit n->infinity [tex]\frac{3}{n^{\frac{n+1}{n}}}[/tex]

    limit n->infinity [tex]\frac{3}{n^1}}[/tex]

    limit ->infinity [tex]0[/tex]

    The series MAY or MAY NOT be convergent.

    Comparison Test
    *Note* This series only contains positive terms*
    From the looks of it, I'm going to GUESS that this series DIVERGES.
    [tex]a_{n} = \frac{3}{n^{1+\frac{1}{n}}}, b_{n} = \frac{1}{n}[/tex]
    [tex]a_{n} \geq b_{n}[/tex]

    Since the series is [tex]\sum^{\infty}_{n=1} \frac{1}{n}[/tex], it is a p-series and it diverges because p [tex]\leq[/tex] 1.

    By the Comparison Test, [tex]\sum^{\infty}_{n=1}\frac{3}{n\sqrt[n]{n}}[/tex] also diverges.

    Limit Comparison Test

    From the looks of it, I'm going to GUESS that this series DIVERGES.
    *Note* This series only contains positive terms*
    [tex]a_{n} = \frac{3}{n^{1+\frac{1}{n}}}, b_{n} = \frac{1}{n}[/tex]

    limit n->infinity [tex]\frac{\frac{3}{n^{\frac{n+1}{n}}}}{\frac{1}{n}}[/tex]

    limit n->infinity [tex]\frac{3n}{n^{\frac{n+1}{n}}}}[/tex]

    limit n->infinity [tex]\frac{3}{n^{\frac{1}{n}}}}[/tex]

    limit n->infinity [tex]0[/tex]


    By the Limit Comparison Test, [tex]\sum^{\infty}_{n=1}\frac{3}{n\sqrt[n]{n}}[/tex] is divergent since 0 > 0.

    My questions:
    Did I pick the right [tex]b_{n}[/tex]? If not, what did I do wrong in picking [tex]b_{n}[/tex]?
    Any hints for picking the proper [tex]b_{n}[/tex]?
    Was there a step that I missed or was unclear?

    As always, any and all help is appreciated and will be greatly thanked! =) (I'm getting a 96% in Calc II thanks to the help I am receiving from this community in understanding concepts! [Nailed a 56/60 on a 20% exam!])

    Sincerely,

    NastyAccident.
     
  2. jcsd
  3. Your bn seems great for your tests.

    [tex]\lim_{n\rightarrow\infty}\frac{3}{n^{\frac{n+1}{n}}}[/tex] does have a limit; look at the denominator n1 + 1/n as nā†’āˆž
     
  4. Mark44

    Staff: Mentor

    For your comparison test, with an being the terms in your series, and bn being the terms in the harmonic series, you said that an >= bn. That very well may be true, but you would need to establish this inequality instead of merely stating it.

    For your work using the limit comparison test, you concluded that
    [tex]\lim_{n \rightarrow \infty}\frac{a_n}{b_n}~=~0[/tex]
    (with an and bn still as defined above), which is not true.

    In your work you show [tex]\frac{\frac{3}{n^{1 + 1/n}}}{\frac{1}{n}}~=~\frac{3}{n^{1/n}}[/tex]
    which is correct, but in evaluating the denominator limit you got an incorrect value. Here's how that goes, looking just at the limit of the denominator:
    [tex]Let~y~=~n^{1/n}[/tex]
    [tex]Then~ln~y~= ln (n^{1/n})~=~1/n*ln~n~=~\frac{ln~n}{n}[/tex]
    Taking the limit of both sides, we have
    [tex]\lim_{n \rightarrow \infty}ln~y~=~\lim_{n \rightarrow \infty}\frac{ln~n}{n}[/tex]
    [tex]=~\lim_{n \rightarrow \infty}\frac{1/n}{1}~=~0[/tex]
    The last limit was evaluated using L'Hopital's Rule.
    Since lim ln y = ln lim y = 0, this means that lim y = 1.

    The upshot of all this is that lim 3/(n1/n) = 3, and not 0 as you wrote.

    This shows that your series diverges, which is in agreement with your instincts.
     
  5. Attempt to satisfy #1 (I'm assuming I do not have to do mathematical induction):
    [tex]a_{n} = \frac{3}{n^{1+\frac{1}{n}}}, b_{n} = \frac{1}{n}[/tex]

    [tex]a_{n} \geq b_{n}[/tex]

    [tex]\frac{3}{n^{1+\frac{1}{n}}}\geq\frac{1}{n}[/tex]

    [tex]\frac{3}{1^{1+\frac{1}{1}}}\geq\frac{1}{1}[/tex]

    [tex]\frac{3}{2^{1+\frac{1}{2}}}\geq\frac{1}{2}[/tex]

    Attempt to satisfy #2 (Please note, I lacked knowledge on Le'Hospital's rule prior to this year since my AB teacher did not teach it.):

    Indeterminate form
    shows that this is infinity to the 0 power case....

    Transformation is:
    [​IMG]

    Which matches what you did.... So =) Thanks for directing me to pull out the Indeterminate forms... I'll be studying them this weekend!

    Sincerely,

    NastyAccident
     
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