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Math Help - [SOLVED] A Series with Partial Sums

  1. #1
    Senior Member Pinkk's Avatar
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    [SOLVED] A Series with Partial Sums

    Investigate the convergence of:

    \sum_{k=1}^{\infty}\,\,\frac{4}{k(k+2)}.

    So I broke it down into:

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

    I have no idea how to write this as the nth partial sum. I tried writing the first few terms:

    S_{3}=2-\frac{2}{3}+1-\frac{1}{2} + \frac{2}{3}-\frac{2}{5}

    I'm completely lost. Thanks!
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  2. #2
    Senior Member Pinkk's Avatar
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    Okay, so looking at it a bit more, and it looks like it'll be S_{n} = 3 - \frac{2}{n+2}

    Is this correct?

    A similar problem

    \sum_{k=1}^{\infty}\,\,\frac{9}{k(k+3)}
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  3. #3
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    Quote Originally Posted by Pinkk View Post
    Okay, so looking at it a bit more, and it looks like it'll be S_{n} = 3 - \frac{2}{n+2}

    Is this correct?
    unfortunately not! the correct answer is: S_n=3 - 2 \left(\frac{1}{n+1} + \frac{1}{n+2} \right). just find S_n for a few values of n and you'll see the pattern!
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  4. #4
    Senior Member Pinkk's Avatar
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    Ah, I think I get it. So using S_{3} in the first problem (since this when terms begin to cancel each other out) you eventually arrive at:

    S_{3}= 3-\frac{2}{4}-\frac{2}{5}=3 - 2(\frac{1}{3+1}+\frac{1}{3+2}) and then simply substitute n=3 in the denominators and I get what you got.

    So for the second problem I posted, is this correct?

    S_{n}=\frac{11}{2} - 3(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}) and thus leading to:

    \sum_{k=1}^{\infty}\,\,\frac{9}{k(k+3)} = \frac{11}{2}
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  5. #5
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    Quote Originally Posted by Pinkk View Post
    Ah, I think I get it. So using S_{3} in the first problem (since this when terms begin to cancel each other out) you eventually arrive at:

    S_{3}= 3-\frac{2}{4}-\frac{2}{5}=3 - 2(\frac{1}{3+1}+\frac{1}{3+2}) and then simply substitute n=3 in the denominators and I get what you got.

    So for the second problem I posted, is this correct?

    S_{n}=\frac{11}{2} - 3(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}) and thus leading to:

    \sum_{k=1}^{\infty}\,\,\frac{9}{k(k+3)} = \frac{11}{2}
    correct! good job!
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  6. #6
    Senior Member Pinkk's Avatar
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    Thanks again. My professor is kinda rushing through this stuff since we're a bit behind schedule, not to mention we have a test on double and triple integration tomorrow, so my brain is on overload right now.
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