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Thread: 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 $\displaystyle x \in \mathbb{R} $ for which the following series converge:

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

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

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

    Attempted solution:
    I thought the only useful test would be the ratio test. So I tried it and got
    $\displaystyle \displaystyle\lim_{n\to\infty} \left|\frac{b_{n + 1}}{b_n}\right| = \displaystyle\lim_{n\to\infty} \frac{2n+ 1}{(n + 1)^3}(x - e)$ however then I found $\displaystyle \displaystyle\lim_{n\to\infty} \frac{2n+ 1}{(n + 1)^3} = 0$ which implies all values $\displaystyle 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...

    $\displaystyle \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 $\displaystyle |x|<1$ ...

    Kind regards

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

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

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

    I think that using the ratio test is simpler.

    $\displaystyle \frac{\frac{x^{n+1} (1-x^{n+1})}{n+1}}{\frac{x^{n} (1-x^{n})}{n}}
    =\frac{x(1-x^{n+1})}{1-x^n}\cdot \frac{n}{n+1}$

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

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

    • If $\displaystyle x=1$,
      $\displaystyle \sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n}=\sum_{n=1}^{\infty} 0.$
      Is that a convergent series ?
    • If $\displaystyle x=-1$,
      $\displaystyle \sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n}=\sum_{n=1}^{\infty} \frac{(-1)^{n}\cdot (1-(-1)^{n})}{n}=\sum_{n=1}^{\infty} \frac{-1}{2n-1}$
      because $\displaystyle 1-(-1)^{n}=0$ when $\displaystyle n$ 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 $\displaystyle x=-1$,
      $\displaystyle \sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n}=\sum_{n=1}^{\infty} \frac{(-1)^{n}\cdot (1-(-1)^{n})}{n}=\sum_{n=1}^{\infty} \frac{-1}{2n-1}$
      because $\displaystyle 1-(-1)^{n}=0$ when $\displaystyle n$ is even. Does this series converge ?
    I believe it is divergent, diverging to $\displaystyle -\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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