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# Sum of a series question

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Hi I need some help with this question.

1. The problem statement, all variables and given/known data

Find the sum of the series.

$\sum_{k=0}^\infty \frac{1}{(k+1)(k+3)}$

3. The attempt at a solution

$= \frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \frac{1}{3\cdot5} + ... + \frac{1}{(n+1)\cdot(n+3)}$

$= \frac{1}{2} [(1-\frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + ... + (\frac{1}{(n+1)} - \frac{1}{(n+3)})$

$= \frac{1}{2}[ 1 + (\frac{1}{2} + \frac{1}{3} + \frac{1}{n+1}) - (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + \frac{1}{n+3}) ]$

So here

$(\frac{1}{2} + \frac{1}{3} + \frac{1}{n+1}) \to 1$

$(\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + \frac{1}{n+3}) \to 0$

Then the whole thing sums to 1?

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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.

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• 1 month later...

I tried like this : $\sum_{k\in\mathbb{N}}\frac{1}{(k+1)(k+3)}=\frac{1}{2}\sum_{k=0}^\infty\frac{1}{k+1}-\frac{1}{k+3}$

$=.5(1+1/2+1/3+1/4+...-1/3-1/4-...)=3/4$ ?

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A23 is absolutely correct! The trick is (which may not be obvious), in

$=.5(1+1/2+1/3+1/4+...-1/3-1/4-...)$

$=.5(1+1/2)=.5(3/2)=3/4$

ALL the remaining terms in the two infinite series cancel out!

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