zeion Posted March 23, 2010 Report Share Posted March 23, 2010 Hi I need some help with this question. 1. The problem statement, all variables and given/known data Find the sum of the series. [math] \sum_{k=0}^\infty \frac{1}{(k+1)(k+3)} [/math] 3. The attempt at a solution[math]= \frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \frac{1}{3\cdot5} + ... + \frac{1}{(n+1)\cdot(n+3)}[/math] [math]= \frac{1}{2} [(1-\frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + ... + (\frac{1}{(n+1)} - \frac{1}{(n+3)})[/math] [math]= \frac{1}{2}[ 1 + (\frac{1}{2} + \frac{1}{3} + \frac{1}{n+1}) - (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + \frac{1}{n+3}) ][/math] So here [math](\frac{1}{2} + \frac{1}{3} + \frac{1}{n+1}) \to 1[/math] [math](\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + \frac{1}{n+3}) \to 0[/math] Then the whole thing sums to 1? Quote Link to comment Share on other sites More sharing options...

lawcat Posted March 23, 2010 Report Share Posted March 23, 2010 I think you'll have to wait for CraigD for this one. But as a preliminary matter, it is a converging series, so a lot of the terms, as they start approaching 4 and 5 decimal places towards zero, you can disregard. You can solve it sufficently by summing up about 8 terms. Of course, the more you go, the better your accuracy. But for a general expression of solution, I can be of no help. Anyway, by inspection, if you use integration for inspection purposes--the area under the curve which essentially sums up all the terms, you should get: 0.5 * ln 3 ~ 0.55. Quote Link to comment Share on other sites More sharing options...

A23 Posted May 14, 2010 Report Share Posted May 14, 2010 I tried like this : [math]\sum_{k\in\mathbb{N}}\frac{1}{(k+1)(k+3)}=\frac{1}{2}\sum_{k=0}^\infty\frac{1}{k+1}-\frac{1}{k+3}[/math][math]=.5(1+1/2+1/3+1/4+...-1/3-1/4-...)=3/4[/math] ? Quote Link to comment Share on other sites More sharing options...

Pyrotex Posted May 14, 2010 Report Share Posted May 14, 2010 A23 is absolutely correct! The trick is (which may not be obvious), in [math]=.5(1+1/2+1/3+1/4+...-1/3-1/4-...)[/math] [math]=.5(1+1/2)=.5(3/2)=3/4[/math] ALL the remaining terms in the two infinite series cancel out! Quote Link to comment Share on other sites More sharing options...

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