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June 4th, 2014, 03:25 AM   #1
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sum2س

sum
1/(1*2*3*4) +4/(2*4*5*6)+9/(5*6*7*8)+..................

Last edited by skipjack; June 4th, 2014 at 06:00 AM.
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June 4th, 2014, 04:34 AM   #2
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$$\frac{n^2}{n(n+1)(n+2)(n+3)}$$

$\displaystyle =\ \frac{-1}{2(n+1)} + \frac2{n+2} + \frac{-3}{2(n+3)}$

$\displaystyle =\ -\frac12\left(\frac1{n+1}-\frac1{n+2}\right) + \frac32\left(\frac1{n+2}-\frac1{n+3}\right)$

Start telescoping.

Last edited by Olinguito; June 4th, 2014 at 04:36 AM.
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June 4th, 2014, 05:19 AM   #3
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Olinguito, your sequence does not match that of the OP when n=2 or n=3. However, I think Mared meant to put a 3 instead of a 2 on the denominator of the second term

$\displaystyle S_n = \frac{1}{1\times2\times3\times4} + \frac{4}{3\times4\times5\times6} + \frac{9}{5\times6\times7\times8} + ...$
$\displaystyle S_n = \sum_1^{\infty}\frac{n^2}{2n(2n-1)(2n+1)(2n+2)} = \sum_1^{\infty}\frac{n^2}{4n(n+1)(2n-1)(2n+1)}$

which can be solved using the same technique that you suggested.
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