Mastering Partial Fractions for Solving Advanced Summation Problems

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embassyhill
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Homework Statement


Handwritten-style math: sum from k=1 to n of 1/((2k-1)(2k+1)) equals 1/(1*3)+1/(3*5)+...+1/((2n-1)(2n+1)), which equals 1/2 times (1-1/3)+1/2 times (1/3-1/5)+...+1/2 times (1/(2n-1)-1/(2n+1)) = 1/2 times (1-1/(2n+1)) = n/(2n+1).

Homework Equations


The Attempt at a Solution


Obviously I don't need a solution because it's right there. What I need to understand is what happened after the third equation sign and more importantly, how would I learn to solve these kinds of problems on my own. I looked at Wikipedia and even though it was helpful, I still don't understand this specific exercise for example. I would be grateful if someone could point me at a resource on this topic.
 
So you're wondering how to go from:

[tex]\frac{1}{2}(\frac{1}{1}-\frac{1}{3}) + \frac{1}{2}(\frac{1}{3}-\frac{3}{5}) + ... + \frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})[/tex]

to:

[tex]\frac{1}{2}(1-\frac{1}{2n+1})[/tex]

Factor a 1/2 out of all the terms and then add the stuff inside. Something interesting happens. For example:

(1/2)(a+b) + (1/2)(c+d) = (1/2)(a+b+c+d)
 
Ok so everything but 1/1 and -1/(2n+1) cancel each other out. Now I get this exercise but what about the bigger question? Are there any tricks to learn here (about the sigma) or is it just about using your knowledge of algebra?
 
It depends on the problem, but a lot of times you can use mathematical induction to prove summation results. There's an example of it on wiki:

http://en.wikipedia.org/wiki/Mathematical_induction

This assumes that you know what you're trying to prove ahead of time (like in the case of your example). However, if you want to know a general way of finding the sum of n terms, there's no standard way, as far as I know. The best you can usually do is some trick like the one I showed you, or see if you're looking at a geometric sum. Otherwise, memorize these:

http://en.wikipedia.org/wiki/Summation
 
Partial fractions:

[tex]\frac{1}{(2x-1)(2x+1)} = \frac{1/2}{2x-1} - \frac{1/2}{2x+1}[/tex]