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How to find sum of series

  1. Nov 28, 2008 #1
    How to find sum of some series?

    For example I got:

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

    All I know is that the condition that I need to find the sum of series is the sum of all partial (separate sums) and I know that the sum must be convergent.

    So [tex]\lim_{n \rightarrow \infty}(x_n)=0[/tex].

    In my case it is true.

    So, [tex]\sum_{n=1}^{\infty}{\frac{1}{n(n+2)}}=\frac{1}{1*3}+\frac{1}{2*4}+\frac{1}{3*5}...[/tex]

    I need to find [tex]\lim_{n \rightarrow \infty}(X_n)[/tex].

    But how will I find the limit of the partial sums?

    Thanks in advance.
     
  2. jcsd
  3. Nov 28, 2008 #2

    mathman

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    Trick: 1/[n(n+2)] = (1/2)(1/n - 1/(n+2)). Separate the sum into sum over odd n and sum over even n. For each sum, the series telescopes to the first positive term. The final result is then (1/2)(1 + 1/2) = 3/4.
     
  4. Nov 28, 2008 #3
    Do you mean like: [tex]\lim_{n \rightarrow \infty}X_n=\lim_{n \rightarrow \infty}[\frac{1}{2}(1- \frac{1}{3})+\frac{1}{2}(\frac{1}{2}-\frac{1}{4})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{4}-\frac{1}{6})][/tex]

    I see that some terms cancel out, but how did you get (1/2)(1 + 1/2) = 3/4 ?
     
  5. Nov 28, 2008 #4

    Office_Shredder

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    You say some terms cancel... look at which ones don't cancel (not many)
     
  6. Nov 29, 2008 #5
    Here cancel 1/3-1/3, 1/4-1/4.
    1/2 * 1 for sure is not canceling, and also 1/2 * 1/2, -1/2 * 1/5, -1/2 * 1/6
    As I can see 1/2*1/n+2 will always stay there. How to get in order these ones?
     
  7. Nov 29, 2008 #6

    mathman

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    I said SEPARATE odd indices from even indices. You have the sum of two sums.

    1/2( 1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 + ....) = 1/2

    and 1/2(1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ....) = 1/4

    Therefore total = 3/4
     
  8. Nov 29, 2008 #7
    Oh, I see. Could you possibly tell me should I always use this method? I mean, should I always use [tex]\frac{x}{n}-\frac{y}{n+2}[/tex] depending from the denominator?

    For example:

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

    should I use [tex]\frac{1}{n^2}-\frac{n^2}{(n+1)^2}[/tex] ?

    Thanks in advance.
     
  9. Nov 30, 2008 #8

    mathman

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    Only if you do it right. (2n+1)/[n2(n+1)2] = 1/n2 - 1/(n+1)2
     
  10. Dec 1, 2008 #9
    Sorry, it was typo mistake.
    [tex]\frac{2n+1}{n^2(n+1)^2}=\frac{A}{(n+1)^2}+\frac{2n+1}{n^2(n+1)^2} - \frac{A}{(n+1)^2}=\frac{A}{(n+1)^2}+\frac{2n+1-An^2}{n^2(n+1)^2}=[/tex]
    n=-1
    [tex]2n+1-An^2=0[/tex]
    A=-1
    [tex]=-\frac{1}{(n+1)^2}+\frac{(n+1)^2}{n^2(n+1)^2}=\frac{1}{n^2}-\frac{1}{(n+1)^2}[/tex]

    Xn=(1 - 1/4) + (1/4 - 1/9) + (1/9 - 1/16) +...+ (1/n2 - 1/(n+1)2)

    [tex]\lim_{n \rightarrow \infty}(X_n)=\lim_{n \rightarrow \infty}(1 - \frac{1}{(n+1)^2})=1-0=1[/tex]

    Thanks for the help and the efforts.
     
    Last edited: Dec 1, 2008
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