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Preliminary Test of Alternating Geometric Series

  1. Apr 2, 2016 #1
    1. The problem statement, all variables and given/known data
    [tex]\lim_{n \to \infty}\frac{(-1)^{n+1} \cdot n^2}{n^2+1}[/tex]

    2. Relevant equations
    [tex]\lim_{n \to \infty}a_n \neq 0 \rightarrow S \ is \ divergent[/tex]

    3. The attempt at a solution
    I tried L'Hopital's rule, but I could not figure out how to find the limit of that pesky [itex](-1)^{n+1}[/itex].

    Edit: This is not actually a geometric series, disregard that part of the title.
     
  2. jcsd
  3. Apr 2, 2016 #2

    Ray Vickson

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    Is there a question somewhere here? You seem to have arrived at the correct conclusion for the correct reason, so what are you unsure about?
     
  4. Apr 3, 2016 #3
    Computing the derivative of an exponential function results in another exponential function, which doesn't really help when it comes to using L'Hopitals rule.
     
  5. Apr 3, 2016 #4

    micromass

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    This is a rule for series. Why is this relevant? There are no series in your post.
     
  6. Apr 3, 2016 #5

    HallsofIvy

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    Your post is very confusing! You title this "alternating geometric series" but appear to be asking about a "sequence" rather than a "series". Further this is not a "geometric" series or sequence. If the "problem statement" is to determine whether or not the series [itex]\sum_{n=0}^\infty \frac{(-1)^n n^2}{n^2+ 1}[/itex] converges then you should know that an alternating series, [itex]\sum_{n=0}^\infty a_n[/itex], converges if and only if the sequence [itex]a_n[/itex] converges to 0. Finally, you ask about the limit of [itex](-1)^n[/itex]. That sequence does not converge so has no limit but that is irrelevant to this problem.
     
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