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

    Now, to telescopic series.

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

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

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

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

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

    In your case:

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

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

    S_n=1-\frac{1}{n+1} and  lim_{n\to\infty}S_n=1, hence \sum_{n=1}^{\infty}(\frac{1}{n(n+1)}=1
    Last edited by Also sprach Zarathustra; March 23rd 2011 at 02:13 PM.
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