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Homework Help: Convergence of a series

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm not sure how to do the notation on here but. Does this series converge or diverge. If it converges, then to what value.
    The series: Sum from 1 to infinity of [(-1)^n * n / (n^2-4n-4)]

    2. Relevant equations
    It tells me to use the ratio test

    3. The attempt at a solution
    I used the ration test and got the limit to equal 1 which is inconclusive. I used the alternating series test to find that it converges, but how do I figure what it converges to and whether it is absolute convergence or not? I'm new to series so this is a little confusing. Any help is appreciated
  2. jcsd
  3. Apr 29, 2012 #2


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    In LaTeX form, this is what your equation appears like:
    [tex]\sum^{\infty}_{n=1} \frac{(-1)^n n}{ (n^2-4n-4)}[/tex]
    Since there is an alternating sign and you need to find the series, i believe you need to use AST.

    The conditions that must be satisfied for the series to converge:[tex]a_n>0
    \\\lim_{n \rightarrow \infty} a_n=0
    \\a_{n+1} \leq a_n[/tex]where[tex]a_n=\frac{n}{n^2-4n-4}[/tex]Indeed, the series converges.
    [tex]\lim_{n \rightarrow {\infty}} \frac{n}{n^2-4n-4}=\lim_{n \rightarrow {\infty}} \frac{1/n}{1-4/n-4/n^2}[/tex]
    I would say it converges to 0?

    To test for absolute or conditional convergence, test if the absolute value of the original series converges or not:
    [tex]\sum^{\infty}_{n=1} \left|\frac{n}{ (n^2-4n-4)}\right|[/tex]
    Last edited: Apr 29, 2012
  4. Apr 29, 2012 #3
    How can you say it converges at 0? When the limit of An is 0 doesn't that just mean that the series converges? Wouldn't I need to write it as a function somehow and take the limit of that to find the convergence? This is an online problem and 0 isn't correct
  5. Apr 29, 2012 #4
    Sorry I was reading the questing wrong. It was asking for my results from the ratio test which explains a lot.
  6. Apr 29, 2012 #5


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    No, it means the series may converge. It may still diverge as well. In any case, what sharks implied above is wrong, as you noted.

    Yes, generally you'll need to find another way to figure out what a series converges to, if it indeed converges.
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