# Telescoping sum

1. Dec 12, 2006

### kasse

1/(16n^2-8n-3)

I want to find the sum of this series from n=1 to infinity

By partial fraction I find

1/(16n^2-8n-3)=1/((4n+3)(4n-1))

which can be written

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

The sum of this is 1/12, but the right answer is 1/4. What is wrong?

Last edited: Dec 12, 2006
2. Dec 12, 2006

### cristo

Staff Emeritus

$$16n^2-8n-3 = (4n+1)(4n-3)$$

3. Dec 12, 2006

### kasse

My mistake

Last edited: Dec 12, 2006
4. Dec 12, 2006

### cristo

Staff Emeritus
The term on the RHS is not the expression you give in your question!

5. Dec 12, 2006

### kasse

I realized that.

The simplest mistakes are often the most difficult to find.