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Alternating Series Test/Test for Divergence

  1. Nov 27, 2012 #1
    So I've been practicing several series that can be solved using the alternating series test, but I've came to a question that's been bothering me for sometime now.

    If a series fails the alternating series test, will the test for divergence always prove it to be divergent?

    Typically, in most examples that I find on the James Stewart website they completely omit the alternating component when using the test for divergence. Is this because it will make the limit not exist?

    Some examples.

    Ʃ n = 1 to ∞ {(-1)^n * 2^(1/n)}

    The alternating series test fails, so in the solution they take the
    lim n → ∞ 2^(1/n) = 1.

    Ʃ n = 1 to ∞ {(-1)^n * ((2^n) / n^2))}

    Similarly, the alternating series test fails, so they use the test for divergence.
    Again, they fail to include the (-1)^n and conclude that

    lim n → ∞ 2^n / n^2 = ∞ ∴ the series is divergent.

    Why do they not include the (-1)^n, wont this make the limit not exist? Obviously, this will still prove the series to diverge, but which one is the correct way to do it? Should I write the limit does not exist or the limit = 1? Thanks in advance to anyone that can help me out!
    Last edited: Nov 27, 2012
  2. jcsd
  3. Nov 27, 2012 #2
    You are right. They need to include the [itex](-1)^n[/itex] in the limit. So you need to find the limit

    [tex]\lim_{n\rightarrow +\infty} (-1)^n 2^{1/n}[/tex]

    which indeed won't exist.

    However, it is obvious that [itex]\lim_{n\rightarrow +\infty} x_n=0[/itex] if and only if [itex]\lim_{n\rightarrow +\infty} |x_n|=0[/itex]. This shows that [tex]\lim_{n\rightarrow +\infty} (-1)^n 2^{1/n}[/tex] is nonzero because [itex]\lim_{n\rightarrow +\infty} 2^{1/n}[/itex] is nonzero.

    So, they are also right. Instead of calculating the limit of the terms of the series, they calculate the limit of the absolute value. And because of that absolute value, the [itex](-1)^n[/itex] factor disappears.
    I do feel that they should mention this explicitely.
  4. Nov 27, 2012 #3
    Thank you. That was helpful : )
  5. Nov 30, 2012 #4
    Just realize that an alternating series is only alternating when it has (-1)^n or (-1)^n+1. So if they posses either of these then you have to figure out what the entire summation is without that part. You then call it (a-sub n) the lim n→∞ of that function. It then has to equal 0 and be greater than (a-sub n+1). If it follows both of these then it will converge, if it doesn't then it diverges.

    With this being said. Your example shows an alternating series, but when you take the lim n→+∞ you get 1. This is not 0 so you immediatly say the series is diverging.
  6. Nov 30, 2012 #5


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    Notice, by the way, that to prove a series, [itex]\sum a_n[/itex] divergent by the "divergence test" it is sufficient to prove that [itex]lim_{n\to\infty} a_n[/itex] is not 0. In fact, if an alternating series does not converge, then that limit does not even exist!
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