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Sum of Infinite series, does it converge?

  1. Jan 16, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the sum of the infinite series, if the series converges.

    n = 1

    (2) / (n^2 + 2n)

    2. Relevant equations


    3. The attempt at a solution

    I believe this problem doesn't look very hard, I think all i really need to do is divide the denominator and numerator by the largest n in the denominator.

    so it would look like this --> (2/n^2) / (1 + 2/n)

    which would become 0/1

    so converges to 0?

    but what is finding the sum of the infinite series?
    help here please. Thankyou.
  2. jcsd
  3. Jan 16, 2007 #2


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    Homework Helper

    You've shown that the terms go to zero as n->infinity. If this isn't true, the series couldn't converge, but by itself it doesn't guarantee the series converges. For example, even though [itex]1/n \rightarrow 0[/itex]:

    [tex] \sum_{n=1}^\infty \frac{1}{n} = \infty [/tex]

    Are you familiar with the series:

    [tex] \sum_{n=1}^\infty \frac{1}{n^2} [/tex]

    ? You can use this to show that your series converges. To find the sum, try to make it into a telescoping series.
  4. Jan 16, 2007 #3
    i'm not too familiar with all the different series quite yet.

    i'm confused when they say 'find the sum'

    Am i trying to prove that this problem is similiar to a series pattern?
    if so, we're on track to use the telescoping series to do this.

  5. Jan 16, 2007 #4


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    Interesting that you should mention "telescoping series"! Use "partial fractions" to write [itex]\frac{2}{n^2+ 2n}[/itex] as difference of two fractions and see what you get.
  6. Jan 16, 2007 #5
    ya someone up above recommended the telescoping series..

    for partial fractions I tried using a formula, but i don't think i reached the correct partial fractions.

    first i factored this fraction: 2/(n)(n+2)

    then used substitution to reach factorials, but i don't think i reached the correct answer --> 2/n(n+2) = 1/n + 1/(n+2)

    are we making progress?
    thanks so far.
  7. Jan 17, 2007 #6


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    It's with a minus

    [tex] \frac{2}{n(n+2)}=\frac{1}{n}-\frac{1}{n+2} [/tex]

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