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Math Help - Values that make series converge

  1. #1
    Junior Member utopiaNow's Avatar
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    Values that make series converge

    Find the values for x \in \mathbb{R} for which the following series converge:

    1. \displaystyle\sum_{n=1}^{\infty}\frac{x^n(1 - x^n)}{n}

    Attempted solution:
    I tried to use the root test to evaluate \displaystyle\lim_{n\to\infty} |a_n|^{1/n} and found that it is equal to  \displaystyle\lim_{n\to\infty} |x(1 - x^n)|^{1/n}. I got stuck here so I tried to use the ratio test and got to \displaystyle\lim_{n\to\infty} |a_n|^{1/n} = \displaystyle\lim_{n\to\infty} \frac{x(1 - x^{n+1})}{1 - x^n}. And I'm stuck here for this method as well.

    2. \displaystyle\sum_{n=1}^{\infty}\left[\frac{(2n)!}{n(n!)^2}\right](x - e)^n

    Attempted solution:
    I thought the only useful test would be the ratio test. So I tried it and got
    \displaystyle\lim_{n\to\infty} \left|\frac{b_{n + 1}}{b_n}\right| = \displaystyle\lim_{n\to\infty} \frac{2n+ 1}{(n + 1)^3}(x - e) however then I found \displaystyle\lim_{n\to\infty} \frac{2n+ 1}{(n + 1)^3} = 0 which implies all values x make this series converge, but that doesn't seem right.

    Any suggestions would be appreciated. Thanks in advance.
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  2. #2
    MHF Contributor chisigma's Avatar
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    The series can be 'splitted' as...

    \sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n} = \sum_{n=1}^{\infty} \frac{x^{n}}{n} - \sum_{n=1}^{\infty} \frac{x^{2n}}{n}

    ... and it becomes the sum of two series both converging for |x|<1 ...

    Kind regards

    \chi \sigma
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by utopiaNow View Post
    2. \displaystyle\sum_{n=1}^{\infty}\left[\frac{(2n)!}{n(n!)^2}\right](x - e)^n

    ...which implies all values x make this series converge, but that doesn't seem right.
    You're correct, it converges for all values of x.
    Quote Originally Posted by chisigma View Post
    The series can be 'splitted' as...

    \sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n} = \sum_{n=1}^{\infty} \frac{x^{n}}{n} - \sum_{n=1}^{\infty} \frac{x^{2n}}{n}

    ... and it becomes the sum of two series both converging for |x|<1 ...
    But it doesn't mean that \sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n} converges iff both \sum_{n=1}^{\infty} \frac{x^{n}}{n} and \sum_{n=1}^{\infty} \frac{x^{2n}}{n} converge. For example for x=1 the first series does converge but the two others don't. It remains to be checked that \sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n} diverges when x=-1 or |x|>1.

    I think that using the ratio test is simpler.

    \frac{\frac{x^{n+1} (1-x^{n+1})}{n+1}}{\frac{x^{n} (1-x^{n})}{n}}<br />
=\frac{x(1-x^{n+1})}{1-x^n}\cdot \frac{n}{n+1}

    If |x|<1, x^n\xrightarrow[n\to\infty]{}0 hence
    \lim_{n\to\infty}\frac{x(1-x^{n+1})}{1-x^n}\frac{n}{n+1}=x\cdot \lim_{n\to\infty}\frac{1-x^{n+1}}{1-x^n}\cdot \lim_{n\to\infty}\frac{n}{n+1}=x\cdot \frac{1-0}{1-0} \cdot 1 =x

    and as |x|<1, the series converges. Similarly one can show that if |x|>1 the series diverges. To check if the series converges when |x|=1 simply substitute x=1 or x=-1 :

    • If x=1,
      \sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n}=\sum_{n=1}^{\infty} 0.
      Is that a convergent series ?
    • If x=-1,
      \sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n}=\sum_{n=1}^{\infty}  \frac{(-1)^{n}\cdot (1-(-1)^{n})}{n}=\sum_{n=1}^{\infty}  \frac{-1}{2n-1}
      because 1-(-1)^{n}=0 when n is even. Does this series converge ?
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  4. #4
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    • If x=-1,
      \sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n}=\sum_{n=1}^{\infty}  \frac{(-1)^{n}\cdot (1-(-1)^{n})}{n}=\sum_{n=1}^{\infty}  \frac{-1}{2n-1}
      because 1-(-1)^{n}=0 when n is even. Does this series converge ?
    I believe it is divergent, diverging to -\infty . Would the formal reason be that it diverges because it is a general harmonic series?
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by utopiaNow View Post
    Would the formal reason be that it diverges because it is a general harmonic series?
    Yes.
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