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Math Help - Discuss the convergence of the series

  1. #1
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    Discuss the convergence of the series

    This question is a weird one. I may have a potential answer for it, but I'm not confident in how good it is.

    Discuss the convergence of the series (as completely as possible)
    \left[\left(1+\frac{1}{2}\right)^2-\left(1+\frac{1}{3}\right)^2\right]+\left[\left(1+\frac{1}{4}\right)^2-\left(1+\frac{1}{5}\right)^2\right]+...+\left[\left(1+\frac{1}{2n}\right)^2-\left(1+\frac{1}{2n+1}\right)^2\right]+...
    What is the effect of removing the square brackets?

    My current answer is uncertain, but here's what I have.

    If the series is \left(1+\frac{1}{2n}\right)^2, we have
    \lim_{n\rightarrow\infty}\left(1+\frac{1}{2n}\righ  t)^2=1
    At a certain point, a_n=\left(1+\frac{1}{2n}\right)^2 is nearly equal to 1.
    As such, adding an infinite number of 1's together cannot be a finite number.
    Therefore, the series is divergent.

    Does my answer have any holes in its logic? And I also haven't figured out what the effect of "removing the square brackets" would do.
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  2. #2
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    Showing that the limit does not cause the expression to go to 0 is called the test for divergence in most calc books, and is an entirely valid method (assuming you have the right series! It should be alternating). The square brackets make it easier to show that this is a telescoping sum, which you can transform into a summation by using partial fractions probably. This is how you should make sure that this is really the series you should be working with!
    Last edited by charmedquark; February 22nd 2011 at 03:18 PM.
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  3. #3
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    Quote Originally Posted by Runty View Post
    Discuss the convergence of the series (as completely as possible)
    \left[\left(1+\frac{1}{2}\right)^2-\left(1+\frac{1}{3}\right)^2\right]+\left[\left(1+\frac{1}{4}\right)^2-\left(1+\frac{1}{5}\right)^2\right]+...+\left[\left(1+\frac{1}{2n}\right)^2-\left(1+\frac{1}{2n+1}\right)^2\right]+...
    What is the effect of removing the square brackets?
    I think that you have really miss read this series.
    It is \displaystyle \sum\limits_{k = 2}^\infty  {\left( { - 1} \right)^k \left( {1 + \frac{1}{k}} \right)^2 }
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  4. #4
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    Quote Originally Posted by Plato View Post
    I think that you have really miss read this series.
    It is \displaystyle \sum\limits_{k = 2}^\infty  {\left( { - 1} \right)^k \left( {1 + \frac{1}{k}} \right)^2 }
    *facepalms* You're right. How did I miss that?

    The series convergent, not divergent! It converges to zero!
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  5. #5
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    Quote Originally Posted by Runty View Post
    The series convergent, not divergent! It converges to zero!
    That is nonsense.
    Does {\left( { - 1} \right)^k \left( {1 + \frac{1}<br />
{k}} \right)^2 }\to 0~?
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  6. #6
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    Quote Originally Posted by Plato View Post
    That is nonsense.
    Does {\left( { - 1} \right)^k \left( {1 + \frac{1}<br />
{k}} \right)^2 }\to 0~?
    Hmm, maybe you're right. I might be taking this too fast.

    Well, that's why I asked in the first place.
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  7. #7
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    Quote Originally Posted by Runty View Post
    Hmm, maybe you're right. I might be taking this too fast.
    Well, that's why I asked in the first place.
    The series diverges by the first test for divergence.
    Do you know what that test says?
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