Telescoping series sum for 1/(16n^2-8n-3)

[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?

 

[cristo]

Well, to start with, your factoring is wrong:

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

 

[kasse]

My mistake

 

[cristo]

$$16n^2-8n-3 = (4n+3)(4n-1)$$ Why isn't this possible? $$(4n+3)(4n-1) = 16n^2-4n+12n-3 = 16n^2+8n-3$$ or what?

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

 

[kasse]

I realized that.

The simplest mistakes are often the most difficult to find.