Finding sum to infinite terms of series:

$$\frac{1}{1\cdot3\cdot5} + \frac{1}{3\cdot5\cdot7} + \frac{1}{5\cdot7\cdot9} + \cdots$$

I approached this question by writing the general term first,then breaking it into partial fractions and then putting the values $1,2,3$ upto $n$ and adding and noticing the terms cancel each other to yield the sum.This method takes a lot of time especially breaking into partial fractions by taking L.C.M. I will be highly thankful if anyone could suggest a faster method.

  • $\begingroup$ what is 1/1.3.5? is it $\frac{1}{1*3*5}=\frac{1}{15}$? $\endgroup$ – Jorge Fernández Hidalgo Dec 4 '13 at 18:32
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    $\begingroup$ $$\frac{1}{(2k+1)(2k+3)(2k+5)} = \frac{1}{4} \left( \frac{1}{(2k+1)(2k+3)} - \frac{1}{(2k+3)(2k+5)}\right)$$ $\endgroup$ – achille hui Dec 4 '13 at 19:01
  • $\begingroup$ Any identities with double factorials you can use? $\endgroup$ – John Dec 4 '13 at 19:05
  • $\begingroup$ Perhaps I should rewrite the hint in this way so that you don't need to remember how to come up with the breakdown. $$\frac{1}{a(a+b)(a+2b)\cdots(a+mb)} = \frac{1}{mb}\left(\frac{(a+mb)-a}{a(a+b)\cdots(a+mb)}\right) \\ = \frac{1}{mb}\left(\frac{1}{a(a+b)\cdots(a+(m-1)b)} - \frac{1}{(a+b)(a+2b)\cdots(a+mb)}\right) $$ $\endgroup$ – achille hui Dec 4 '13 at 19:16

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