# Telescoping Series

1. Mar 18, 2005

### 7C0A0A5

I have no idea how to do this :'( really the only part I don't understand is the ending part.....like for

the infinite sum of (1/n(n+1)).............I know you start of by partial fractions.....then you just plug in a few numbers for n.

so I end up with:

(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1))

then in another problem...the infinite sum of ( 2/{(n-1)(n+1)} ) ends up like...

(1 - 1/3) + (1/2 - 1/4) +1/3 - 1/5) + ... + (1/(n-3) - 1/(n-1)) + (1/(n-2) - 1/n)

I want to know how you get the end results.....the ones with "n" in them....I don't understand how to get those numbers....or why they are what they are....

I understand the first part without the "n"....but I don't know how to end it with the "n"....I hope this makes since.

2. Mar 18, 2005

### NateTG

OK, let's take a look at:
$$\sum_{n=1}^{4}\frac{1}{n}-\frac{1}{n+1}$$
So if we write it out longhand it turns into:
$$(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})$$
but we can regroup:
$$\frac{1}{1}+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+(-\frac{1}{4}+\frac{1}{4})-\frac{1}{5}$$
This is just moving the parens around, but the paired numbers nicely all zero out, leaving you with
$$\frac{1}{1}-\frac{1}{5}$$

It should be easy to see that the pattern can be extended to any number of terms so that you end up with
$$\sum_{n=1}^{k}\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{k+1}$$

There are more formal approaches to this, but I doubt that they will be particularly helpful to you.

Last edited: Mar 18, 2005