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Thread: telescopeing series?

  1. #1
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    telescopeing series?

    i dont understand how i tell if a telescopeing series is convergent or divergent...

    problem:
    1/n(n+1)

    i used partial fractions and got it down to this....

    S_n = 1 - 1/(n+1) = total sum = 1

    what do i look at/do to know if this is convergent or divergent
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by lochnessmonster View Post
    i dont understand how i tell if a telescopeing series is convergent or divergent...

    problem:
    1/n(n+1)

    i used partial fractions and got it down to this....

    S_n = 1 - 1/(n+1) = total sum = 1

    what do i look at/do to know if this is convergent or divergent

    The series $\displaystyle \sum_{k=1}^{\infty}a_k$ is converges if there exist a finite limit $\displaystyle S$ of partial sum sequence $\displaystyle \{S_n\}$ and $\displaystyle S$ is the sum of the series $\displaystyle \sum_{k=1}^{\infty}a_k$.
    If the limit $\displaystyle \{S_n\}$ isn't exist, we say that the series is divergent.

    Now, to telescopic series.

    $\displaystyle \sum_{n=1}^{\infty}(a_{n+1}-a_n)$ is telescopic series.

    Partial sum $\displaystyle S_n$ of $\displaystyle \sum_{n=1}^{\infty}(a_{n+1}-a_n)$ is:

    $\displaystyle S_n=(a_2-a_1)+(a_3-a_2)+...+(a_{n+1}-a_n)=a_{n+1}-a_1$

    By the definition above $\displaystyle \sum_{n=1}^{\infty}(a_{n+1}-a_n)$ converges if $\displaystyle lim_{n\to\infty}S_n$ exist.

    Hence, $\displaystyle lim_{n\to\infty}S_n=lim_{n\to\infty}a_{n+1}-a_1$, so $\displaystyle \sum_{n=1}^{\infty}(a_{n+1}-a_n)$ converges iff $\displaystyle \{a_n\}$ converges.

    In your case:

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

    $\displaystyle \{\frac{1}{n}\}$ is converges sequence, so the series is also converges.

    $\displaystyle S_n=1-\frac{1}{n+1}$ and $\displaystyle lim_{n\to\infty}S_n=1$, hence $\displaystyle \sum_{n=1}^{\infty}(\frac{1}{n(n+1)}=1$
    Last edited by Also sprach Zarathustra; Mar 23rd 2011 at 01:13 PM.
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