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Math Help - Sum and convergence of Series

  1. #1
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    Sum and convergence of Series

    Hey,

    I need some help finding the sum of this and showing it converges. To show a series is convergent do you show its limit approaches zero? So for a divergent series you would show it doesn't approach zero.

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

    Thanks
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  2. #2
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Benno_11 View Post
    Hey,

    I need some help finding the sum of this and showing it converges. To show a series is convergent do you show its limit approaches zero? So for a divergent series you would show it doesn't approach zero.

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

    Thanks
    No, the test is inconclusive when the limit of the summand is zero (eg. the series can converge or diverge). Try the ratio test.
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  3. #3
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    Telescoping sum

    Use partial fractions.

    \sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2} = \sum_{n=1}^\infty \frac{A}{n^2}+ \frac{B}{(n+1)^2} for some integers A,B

    \sum_{n=1}^\infty \frac{A}{n^2}+ \frac{B}{(n+1)^2} = \sum_{n=1}^\infty \frac{A(n+1)^2+Bn^2}{n^2(n+1)^2} = \sum_{n=1}^\infty \frac{(A+B)n^2+2An+1}{n^2(n+1)^2}

    So A+B=0 and 2A=2, therefore A=1 and B=-1, so

    \sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2} = \sum_{n=1}^\infty \frac{1}{n^2}- \frac{1}{(n+1)^2} = \frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16} +...

    Notice all but the last terms cancel each other, so...

    \sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2} = \lim_{L\rightarrow\infty} 1-\frac{1}{L^2}=1

    QED
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  4. #4
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    Quote Originally Posted by Benno_11 View Post
    Hey,

    I need some help finding the sum of this and showing it converges. To show a series is convergent do you show its limit approaches zero? no

    So for a divergent series you would show it doesn't approach zero.

    if the nth term does not approach zero, the series diverges ... but note that there are series whose nth terms do approach zero and diverge.

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

    Thanks
    the series is a telescoping series. using the method of partial fractions ...

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

    list out a few terms and a pattern emerges, note that the inner terms cancel ...

    \left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{9}\right) + \left(\frac{1}{9} - \frac{1}{16}\right) + ... + \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) + ... = \lim_{n \to \infty} 1 - \frac{1}{(n+1)^2} = 1
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  5. #5
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    convergence is pretty easy to prove since for each n\ge1 is \frac{2n+1}{n^{2}(n+1)^{2}}< \frac{3n}{n^{4}}=\frac{3}{n^{3}}, thus \sum\limits_{n=1}^{\infty }{\frac{2n+1}{n^{2}(n+1)^{2}}}< \sum\limits_{n=1}^{\infty }{\frac{3}{n^{3}}}<\infty . (Direct comparison test with a p-series.)

    now, to split the original fraction into two ratios just use algebra!

    \frac{2n+1}{n^{2}(n+1)^{2}}=\frac{\left( n^{2}+2n+1 \right)-n^{2}}{n^{2}(n+1)^{2}}=\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}, which is clearly a telescoping series as skeeter stated above.
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