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Math Help - discuss the convergence of the series

  1. #1
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    discuss the convergence of the series

    [(1+\displaystyle \frac{1}{2})^2- (1+\displaystyle \frac{1}{3})^2]+[(1+\displaystyle \frac{1}{4})^2-(1+\displaystyle \frac{1}{5})^2]+...+[(1+\displaystyle \frac{1}{2n})^2 - (1+\displaystyle \frac{1}{2n+1})^2]+...

    What is the effect of removing the square brackets?


    does removing the square brackets mean remove ()^2?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by wopashui View Post
    [(1+\displaystyle \frac{1}{2})^2- (1+\displaystyle \frac{1}{3})^2]+[(1+\displaystyle \frac{1}{4})^2-(1+\displaystyle \frac{1}{5})^2]+...+[(1+\displaystyle \frac{1}{2n})^2 - (1+\displaystyle \frac{1}{2n+1})^2]+...

    What is the effect of removing the square brackets?


    does removing the square brackets mean remove ()^2?
    Removing the square brackets means that the general terms (-1)^{n} (1+\frac{1}{n})^{2} doesn't tend to 0 if n tends to infinity...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by chisigma View Post
    Removing the square brackets means that the general terms (-1)^{n} (1+\frac{1}{n})^{2} doesn't tend to 0 if n tends to infinity...

    Kind regards

    \chi \sigma

    so does the series converge to 0 at the beginning? then diverge after removing the bracket?
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  4. #4
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    Quote Originally Posted by wopashui View Post
    so does the series converge to 0 at the beginning? then diverge after removing the bracket?
    Do you agree that \displaystyle\sum\limits_{k = 2}^\infty  {( - 1)^k \left( {1 + \frac{1}{k}} \right)^2 } is the same as you posted?

    If not, why not?

    If yes, for any series \sum\limits_{n = k}^\infty  {a_n } if (a_n)\not\to 0 then the series diverges.
    This is known as the first divergent test.
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