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Determining the limit of a sequence

  1. Feb 20, 2006 #1
    I have the following sequence

    [tex]

    \begin{array}{l}
    a_n = ( - 1)^n \left( {\frac{n}{{n + 1}}} \right) \\
    \\
    \mathop {\lim }\limits_{n \to \infty } \left[ {( - 1)^n \left( {\frac{n}{{n + 1}}} \right)} \right] \\
    \end{array}



    [/tex]

    Direct substitution yields [tex]
    ( - 1)^\infty \left( {\frac{\infty }{\infty }} \right)
    [/tex]

    I tried manipulating it into a form in which I could apply L'Hopital's Rule.

    [tex]


    \displaylines{
    {\rm Let y} = \mathop {\lim }\limits_{n \to \infty } \left[ {( - 1)^n \left( {\frac{n}{{n + 1}}} \right)} \right] \cr
    \cr
    \ln y = \mathop {\lim }\limits_{n \to \infty } \ln \left[ {( - 1)^n \left( {\frac{n}{{n + 1}}} \right)} \right] \cr
    \cr
    = \mathop {\lim }\limits_{n \to \infty } \left[ {\ln ( - 1)^n + \ln (n) - \ln (n + 1)} \right] \cr
    \cr
    = \mathop {\lim }\limits_{n \to \infty } \left[ {n\ln ( - 1) + \ln (n) - \ln (n + 1)} \right] \cr
    \cr
    \ln ( - 1) = undefined \cr}


    [/tex]

    The answer is below. How did the book arrive at that answer? How did they go through and calculate the limit? Solutions manuals are so wonderfully detailed :)

    [​IMG]
     
    Last edited: Feb 20, 2006
  2. jcsd
  3. Feb 20, 2006 #2
    [tex](\frac{n}{n+1})=(1-\frac{1}{n+1})[/tex]
    hope that helps
     
  4. Feb 20, 2006 #3
    Are you required to test the series this way? I would suggest looking at [tex] \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} [/tex] in which case that nasty power of n cancels out. If the limit above approaches -1, then the series limit is undetermined. (It may converge, it may not.)

    -Dan
     
  5. Feb 20, 2006 #4

    benorin

    User Avatar
    Homework Helper

    It is a sequence opposed to a series that is being tested for convergence, and hence only [tex]\lim_{n\rightarrow\infty}a_n[/tex] need be considered (and for such happyg1's note ought be sufficient.)
     
  6. Feb 21, 2006 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Another way to look at it is to divide both numerator and denominator by n:
    [tex](-1)^n\frac{n}{n+1}= (-1)^n\frac{1}{1+\frac{1}{n}}[/tex]

    As n-> infinity, 1/n-> 0.
     
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