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Integral on sequence

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data
    show that holds,

    [itex] \sum\limits_{n=1}^{\infty}\int\limits_{0}^{\frac{\pi}{2}}\frac{(2n-1) sin(2n-1) x}{n^2(n+1)}dx = \sum\limits_{n=2}^{\infty}\frac{1}{n^2}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    Actually, I have no idea how should I start.
     
  2. jcsd
  3. Sep 10, 2011 #2

    micromass

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    This reduces to

    [tex]\sum_{n=1}^{+\infty}{\frac{2n-1}{n^2(n+1)}\int_0^{\pi/2}{\sin((2n-1)x)dx}}[/tex]

    Now, calculate the integral.
     
  4. Sep 10, 2011 #3
    thank you very much for your help, I calculate the integral;

    [itex]= \sum\limits_{n=1}^{\infty}\frac{(2n-1)}{n^2(n+1)}\Big( \frac{-1}{(2n-1)} \cos((2n-1)x) \vert_{0}^{\frac{\Pi}{2}} \Big)[/itex]

    [itex]= \sum\limits_{n=1}^{\infty}\frac{-1}{n^2(n+1)} \Big( cos((2n-1)\frac{\Pi}{2}) -1 \Big)[/itex]

    [itex]= \sum\limits_{n=1}^{\infty} \frac{1- cos((2n-1)\frac{\Pi}{2}) }{n^2(n+1)}[/itex]

    for [itex]n=1[/itex], this summation becomes [itex]\frac{1}{2}[/itex]

    so,

    [itex]\frac{1}{2} +\sum\limits_{n=2}^{\infty} \frac{1- cos((2n-1)\frac{\Pi}{2}) }{n^2(n+1)}[/itex]

    it becomes like that. but here how should I continue to show that this equation equal to

    [itex] \sum\limits_{n=2}^{\infty} \frac{1}{n^2}[/itex]

    ???????
     
  5. Sep 10, 2011 #4

    micromass

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    OK, but [itex]\cos((2n-1)\pi/2)[/itex] can be calculated, right??

    I have no idea why you drop the summation here...
     
  6. Sep 10, 2011 #5
    Ok if I rewrite it, finally it becomes
    [itex]\frac{}{}[/itex]
    [itex]\[=\sum\limits_{n=1}^{\infty}\frac{1- \cos((2n-1)\frac{\Pi}{2})}{n^2(n+1)}\][/itex]
    [itex]\[=\sum\limits_{n=1}^{\infty}\frac{1- \cos(n \Pi - \frac{\Pi}{2})}{n^2(n+1)}\][/itex]
    [itex]\[=\sum\limits_{n=1}^{\infty}\frac{1- \Big(\cos(n\Pi)\cos(\frac{\Pi}{2}) + \sin(n\Pi)\sin(\frac{\Pi}{2})\Big)}{n^2(n+1)}\][/itex]
    [itex]\[=\sum\limits_{n=1}^{\infty}\frac{1}{n^2(n+1)}\][/itex]

    actually I can't go on ..

    how it is equal to
    [itex]\[=\sum\limits_{n=2}^{\infty}\frac{1}{n^2}\][/itex]
    ????
     
  7. Sep 10, 2011 #6

    micromass

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    OK, now split

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

    into partial fractions...
     
  8. Sep 10, 2011 #7
    ok now I think we found, can you check,

    [itex]=\sum\limits_{n=1}^{\infty}\frac{1}{n^2(n+1)}[/itex]
    [itex]=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - \sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)}[/itex]
    [itex]=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - \Big( \sum\limits_{n=1}^{\infty}\frac{1}{n}- \sum\limits_{n=1}^{\infty}\frac{1}{(n+1)} \Big)[/itex]
    [itex]=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - 1[/itex]
    [itex]=\sum\limits_{n=2}^{\infty}\frac{1}{n^2}[/itex]

    thank you very much....
     
  9. Sep 10, 2011 #8

    micromass

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    Nonono, you can't write that. [itex]\sum_{n=1}^{+\infty}{\frac{1}{n}}[/itex] is infinity. So what's standing there is infinity - infinity. This doesn't make any sense.

    You'll need another way to calculate

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

    Write out the first 10 terms of the sum and see if you notice anything...
     
  10. Sep 10, 2011 #9
    [itex]=\sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)}[/itex]
    [itex]=\sum\limits_{n=1}^{\infty}\Big( \frac{1}{n} - \frac{1}{n+1} \Big)[/itex]
    [itex]= \Big( \frac{1}{1} - \frac{1}{2} \Big)[/itex]
    [itex]~~+ \Big( \frac{1}{2} - \frac{1}{3} \Big)[/itex]
    [itex]~~+ \Big( \frac{1}{3} - \frac{1}{4} \Big)[/itex]
    [itex]~~+ \Big( \frac{1}{4} - \frac{1}{5} \Big)[/itex]
    ........
    and so on,
    then it is equal to [itex] 1 [/itex]
    ..
     
  11. Sep 10, 2011 #10

    micromass

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    Yes, that is the correct reasoning!!
     
  12. Sep 10, 2011 #11
    Thank you very much..
     
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