Closed form for a sequence

1. Jan 29, 2013

bonfire09

1. The problem statement, all variables and given/known data
Let Ʃ 1\(n^2-1) from n=2 to k. It says find a closed for it and prove it using sum notation.

2. Relevant equations

3. The attempt at a solution
I can easily prove it by induction but I don't know what a closed form means. I tried looking it up online but there really isn't much info and nothing is stated in my textbook about. All I know is the sequence when expanded looks like

1/3+1/8+1/15+1/24+...+1/n^2-1. Then not sure how to put it in closed form?

2. Jan 29, 2013

Dick

Closed form means find a formula for the sum of series without writing out each term. Try to express 1/(n^2-1) in the form A/(n+1)+B/(n-1). Find A and B. Like partial fractions. Then start writing out terms of that and think about it.

Last edited: Jan 29, 2013
3. Jan 30, 2013

bonfire09

This is what I get$$\left( \sum_{n=0}^k\frac{1}{-2(n+1)}+\frac{1}{2(n-1)} \right)=\frac{1}{3}+\frac{1}{8}+...+\frac{1}{-2(k+1)}+(\frac{1}{2(k-1)})$$. From here im not sure how to get a closed sum.

4. Jan 30, 2013

Dick

Yes! It's (1/2)*(1/(n-1)-1/(n+1)). Start the sum at n=2 like your original problem posed and bring the (1/2) outside. The series then goes like 1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7... Don't you see a lot of cancellations in there?

5. Jan 30, 2013

bonfire09

Yes but all I'm left is 1+1\2?

6. Jan 30, 2013

Dick

If you are doing an infinite sum, yes. But if you dealing with the finite sum 2...k you also have some stuff that won't cancel at the k end of the sum.