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Math Help - nth partial sum

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    nth partial sum

    Find a formula for the nth partial sum of the series and use it to find the seriesʹ sum if the series converges.

    1)



    I have no idea how to do this one. I missed it on the test and it might show up on the final
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    MHF Contributor chisigma's Avatar
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    The 'key' is the identity...

    \frac{1}{(n+1)(n+2)}= \frac{1}{n+1} - \frac{1}{n+2} (1)

    Kind regards

    \chi \sigma
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    i'm still not getting it
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    MHF Contributor harish21's Avatar
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    Quote Originally Posted by krzyrice View Post
    Find a formula for the nth partial sum of the series and use it to find the seriesʹ sum if the series converges.

    1)



    I have no idea how to do this one. I missed it on the test and it might show up on the final
    \frac{7}{(n+1)(n+2)} = \frac{7}{n+1}-\frac{7}{n+2}

    Now,

    \sum_{n=1}^k  \frac{7}{n+1}-\frac{7}{n+2} =  \left( \frac{7}{2}-\frac{7}{3} \right) + \left( \frac{7}{3}-\frac{7}{4} \right)+....+ \left( \frac{7}{k+1}-\frac{7}{k+2}\right)

    = \left( \frac{7}{2} -\frac{7}{k+2}\right)

    find the limit as k tends to infinity.
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    thanks makes sense now
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  6. #6
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    Quote Originally Posted by krzyrice View Post
    Find a formula for the nth partial sum of the series and use it to find the seriesʹ sum if the series converges.

    1)



    I have no idea how to do this one. I missed it on the test and it might show up on the final
    Surely this is

    7\sum_{n = 2}^{\infty}\frac{1}{n(n + 1)}.


    Using partial fractions:

    \frac{A}{n} + \frac{B}{n + 1} = \frac{1}{n(n + 1)}

    \frac{A(n + 1) + Bn}{n(n + 1)} = \frac{1}{n(n + 1)}

    A(n + 1) + Bn = 1

    (A + B)n + A= 0n + 1.

    So A = 1 and B = -1.


    Therefore we can write the sum as

    7\sum_{n = 2}^{\infty}\frac{1}{n(n + 1)} = 7\sum_{n = 2}^{\infty}\left(\frac{1}{n} - \frac{1}{n + 1}\right)

    This is a telescoping series, so all terms except the first and last will cancel.

    So the n^{\textrm{th}} partial sum is

    7\left(\frac{1}{2} - \frac{1}{n + 1}\right).


    Therefore, the sum of the series is

    \lim_{n \to \infty}7\left(\frac{1}{2} - \frac{1}{n + 1}\right)

     = 7\left(\frac{1}{2} - 0\right)

     = \frac{7}{2}.
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