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Comparison test for convergence

  1. Dec 9, 2007 #1

    tony873004

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    Science Advisor
    Gold Member

    1. The problem statement, all variables and given/known data
    [tex]\sum\limits_{n = 2}^\infty {\frac{{\sqrt n }}{{n - 1}}} [/tex]


    2. Relevant equations



    3. The attempt at a solution

    [tex]\begin{array}{l}
    a_n = \frac{{\sqrt n }}{{n - 1}},\,\,b_n = \frac{{\sqrt n }}{n} = \frac{{n^{1/2} }}{n} = \frac{1}{{n^{1/2} }}\,\,{\rm{is_ a_ p - series_ with }}p{\rm{ = }}\frac{1}{2},\,\frac{1}{2} < 1\,{\rm{diverges}} \\
    \\
    \frac{{\sqrt n }}{{n - 1}} > \frac{{\sqrt n }}{n}\, \\
    \\
    \end{array}[/tex]
    so it is divergent.

    But in this example from the notes, the teacher used the Limit Comparison Test, taking the [tex]\mathop {\lim }\limits_{n \to \infty } \frac{{a_n }}{{b_n }}[/tex], and using this to justify that since bnis is divergent, therefore anis also divergent. I understand this logic, but why was the extra step necessary when the comparison test alone showed it to be divergent?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
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