Understanding telescoping series

1. Nov 13, 2008

jinksys

I understand that if given the following series:

$$\sum _{n=1}^{\infty } \frac{1}{n(n+3)}$$

I can break it up using partial fraction decomposition into:

$$\sum _{n=1}^{\infty } \frac{1}{3n}-\frac{1}{3n+9}$$

If I start listing the values of $$n_1,n_2$$, etc, I can see the cancellation pattern and find the sum.

This process seems tedious, is there another way of doing these problems?

Wikipedia says:

* Let k be a positive integer. Then

$$\sum^\infty_{n=1} {\frac{1}{n(n+k)}} = \frac{H_k}{k}$$

where Hk is the kth harmonic number. All of the terms after 1/(k − 1) cancel.

But I'm not sure what the 1/(k - 1) means, nor do I understand the harmonic number bit either.

2. Nov 13, 2008

Pacopag

I think that what you are doing is correct, and probably the intended method to use. Sometimes you just gotta go ahead and brute force your way through it.

3. Nov 13, 2008

jinksys

Ok,

Do you know what my wikipedia snippet is referring to?
http://en.wikipedia.org/wiki/Telescoping_series

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