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Determine if the series is convergent or divergent

  1. Aug 2, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    [itex]1+\dfrac{2^2}{2!}+\dfrac{3^2}{3!}....... \infty[/itex]


    3. The attempt at a solution

    [itex]t_n = \dfrac{n^2}{n!} \\ \dfrac{n}{(n-1)(n-2).....1}[/itex]

    I tried applying the Ratio Test but couldn't find another function which would give me a finite limit when divided by that function.
     
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  3. Aug 2, 2014 #2

    ehild

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  4. Aug 2, 2014 #3

    vanhees71

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    Well, the ratio test is the following. Given a (formal) series
    [tex]S=\sum_{n=0}^{\infty} a_n[/tex]
    with [itex]a_n \geq 0[/itex], you compare it to the geometric series
    [tex]S'=\sum_{n=0}^{\infty} q^n, \quad q \geq 0[/tex]
    The geometric series is convergent for [itex]0 \leq q<1[/itex].

    This implies that if for some [itex]N[/itex] you have [itex]a_n<q^n[/itex] for all [itex]n>N[/itex] with some [itex]0 \leq q<1[/itex], the series is convergent. No it's easy to show that this is the case if
    [tex]\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}<1.[/tex]
    Note that this makes, of course, only sense if the ratio converges. Also it's important to keep in mind that this ratio cirterion is only suffcient for convergence of the series but by no means necessary, i.e., even if the ratio converges to 1, it doesn't mean that the series is necessarily divergent. For sure, it is divergent, if the ratio converges to a limit [itex]>1[/itex].

    If I say more about the current question, I'd solve the task, but this the OP should be able to do now ;-).
     
  5. Aug 2, 2014 #4

    utkarshakash

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    The above link describes the D' Alembert's Ratio Test which is different from my approach. I should have rather used the word 'Comparison Test' to make things more clear. I'm trying to use the following property:

    If two positive term series ∑un and ∑vn be such that
    [itex]\lim_{n \to \infty} \dfrac{u_n}{v_n} [/itex] is a finite quantity (≠0), then ∑un and ∑vn converge or diverge together.
     
  6. Aug 2, 2014 #5

    ehild

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    I see. But applying the Ratio Test solves your problem about the convergence.

    ehild
     
  7. Aug 2, 2014 #6

    pasmith

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    Use the ratio test, which has the advantage of not requiring you to find a series [itex]\sum a_n[/itex] with known convergence properties and satisfying [itex]\lim_{n \to \infty} a_n n!/n^2 \neq 0[/itex].
     
  8. Aug 2, 2014 #7

    Ray Vickson

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    If all you want to do is prove (or disprove) convergence, just use the easiest tool available---the ratio test in this case. There is no good reason to use the comparison test unless your teacher has ordered you to use it and told you that you are not allowed to use any other method.
     
    Last edited: Aug 2, 2014
  9. Aug 2, 2014 #8

    ehild

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    The Comparison test means if 0<an<bn and ∑bn converges so does ∑an. What you want to apply is the Limit comparison test.

    http://en.wikipedia.org/wiki/Convergent_series

    ehild
     
  10. Aug 2, 2014 #9

    LCKurtz

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    So, Utkarshakash, have you actually tried the ratio test (not the limit comparison test)? What happens?
     
  11. Aug 2, 2014 #10

    utkarshakash

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    I'm not allowed to use the ratio test for this problem.
     
  12. Aug 2, 2014 #11

    SammyS

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    In post #8, ehild pointed out that the test you describe is known as the Limit Comparison Test.


    Try expressing the numerator as ##\ (n-1)+1\ ## and split this result for tn into two terms
     
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