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Math Help - series..diverging, converging

  1. #1
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    series..diverging, converging

    Q2004.
    Can you help me with this question..
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by champrock View Post
    Q2004.
    Can you help me with this question..
    Note that \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}

    so this telescopes to 1. So it is none of these.
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    This seems like a good candidate for "plug and chug". Just try a few values and see what happens. You'll figure it out.
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    Quote Originally Posted by TheEmptySet View Post
    Note that \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}

    so this telescopes to 1. So it is none of these.
    i has used your method and found the value to be 1-Lim(1/n+1) where n tends to infinity.

    So this becomes 1-(some minutest number close to 0) . so shouldnt the answer be "some number between 0 and 1" ?

    i am confused on this part.
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    Quote Originally Posted by champrock View Post
    i has used your method and found the value to be 1-Lim(1/n+1) where n tends to infinity.

    So this becomes 1-(some minutest number close to 0) . so shouldnt the answer be "some number between 0 and 1" ?

    i am confused on this part.
    \sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\sum_{n=1}^{\i  nfty}\frac{1}{n}-\frac{1}{n+1}=\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\frac{1}{n+1}=

    Now lets take out the first term of the series

    1+\sum_{n=2}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\frac{1}{n+1}

    Now lets reindex the first sum with k=n-1 \iff n=k+1

    Note: dont forget that the index is a "dummy variable"

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

    Since the index is a dummy vairable we can combine the sums let k=n

    1+\sum_{n=1}^{\infty}\frac{1}{n+1}-\sum_{n=1}^{\infty}\frac{1}{n+1}=1+\sum_{n=1}^{\in  fty}\left(\frac{1}{n+1}-\frac{1}{n-1} \right)=1+\sum_{n=1}^{\infty}0=1
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