What is a Closed Form for the Sequence?

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bonfire09
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Homework Statement


Let Ʃ 1\(n^2-1) from n=2 to k. It says find a closed for it and prove it using sum notation.

Homework Equations





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?
 
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bonfire09 said:

Homework Statement


Let Ʃ 1\(n^2-1) from n=2 to k. It says find a closed for it and prove it using sum notation.

Homework Equations


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?

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.
 
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This is what I get[tex]\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)})[/tex]. From here I am not sure how to get a closed sum.
 
bonfire09 said:
This is what I get[tex]\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)})[/tex]. From here I am not sure how to get a closed sum.

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?
 
Yes but all I'm left is 1+1\2?
 
bonfire09 said:
Yes but all I'm left is 1+1\2?

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.