How Can You Find the Sum of a Convergent Series with Partial Fractions?

[BuBbLeS01]

Homework Statement

Find the sum of the convergent series:
The sum of 1/ (n^2 - 1) from n=2 to infinity

Homework Equations

The Attempt at a Solution

I want to break it down into 2 fractions and use partial fractions.

1/(n-1)(n+1)...but I don't know where to go from here...

 

[HallsofIvy]

Homework Statement

Find the sum of the convergent series:
The sum of 1/ (n^2 - 1) from n=2 to infinity

Homework Equations

The Attempt at a Solution

I want to break it down into 2 fractions and use partial fractions.

1/(n-1)(n+1)...but I don't know where to go from here...

How about doing what you just said: use partial fractions. If
$$\frac{1}{n^2-1}= \frac{A}{n-1}+ \frac{B}{n+1}$$
what are A and B?

 

[BuBbLeS01]

A would be 1 * (n+1) = n+1
B would be 1 * (n-1) = n-1
right? or no??

 

[Dick]

Noo. Take what Halls gave you and multiply both sides by (n-1)*(n+1) (remembering (n-1)*(n+1)=n^2-1). That should give you an equation to solve for A and B. What is it?

 

[BuBbLeS01]

So I would have
A(n+1) * B(n-1)

 

[Dick]

No. That's not even an equation. It should 'equal' something. And how did you get the '*' between the two terms. Try again. Show more work if you are really confused.

 

[BuBbLeS01]

Ok that should have been + not * my mistake...
1/n^2 - 1 = A/n-1 + B/n+1
1/n^2 - 1 = A(n-1)(n+1)/n-1 + B(n-1)(n+1)/n+1
1/n^2 - 1 = A(n+1) + B(n-1)

 

[Dick]

You are getting closer, but you didn't multiply the left side by (n^2-1), did you?

 

[BuBbLeS01]

Oh I didn't realize I had to...
So I get...
1 = A(n+1) + B(n-1)

 

[Dick]

You've got it. Now write that as 1=(A+B)*n+(A-B). There's no 'n' on the left side so the coefficient of n on the right side should be zero, and the constant term should be 1. Can you find A and B?

 

[BuBbLeS01]

Can I also just do...
1 = A(n+1) + B(n-1)
Plug in 1 to solve for A and -1 to solve for B? Getting 1/2 and -1/2?

 

[Dick]

Sure. Plugging in values of n will work too. But plugging in just n=1 doesn't tell you what B is, does it?

 

[seroth]

can't he plug in -1 for to zero out A and get B

(heaviside cover up method?)

 

[BuBbLeS01]

Yea that's what I meant 1 to get A and -1 to get B

 

[Dick]

I said, sure you can! You could also, without paying much attention to n, just say (A+B)=0, and (A-B)=1 and solve them simultaneously. Whichever method you prefer. They both work. Now can you sum the series?

 

[BuBbLeS01]

I said, sure you can! You could also, without paying much attention to n, just say (A+B)=0, and (A-B)=1 and solve them simultaneously. Whichever method you prefer. They both work. Now can you sum the series?

1/n^2 - 1 = .5/n-1 - .5/n+1

Then I start with plugging in 2 going to infinity...

[.5/2-1 - .5/2+1] + [.5/3-1 - .5/3+1] + [.5/4-1 - .5/4+1] + [.5/5-1 - .5/5+1]...and so on...

[.5/1 - .5/3] + [.5/2 - .5/4] + [.5/3 - .5/5] + [.5/4 - .5/6]...

Some cancel out...I thought it was supposed to be all of them cancel except the 1st and nth term? Somehow I didn't get that...

 

[Dick]

The next term that you didn't write will cancel the .5/5. The one after that will cancel the .5/6. The one after that will cancel the .5/7. Etc. In the end, which terms don't cancel?

 

[BuBbLeS01]

I thought it was supposed to be the first and last but then .5/2 didn't cancel which is half of the 2nd term?

 

[Dick]

I'm not sure why you are saying 'first' and 'last' should cancel, but you are right. .5/2 doesn't cancel.

 

[Feldoh]

Strictly speaking a telescoping series doesn't have to be just the first and last term, as long as enough terms cancel then that's all that matters.

 

[BuBbLeS01]

so as n approaches infinity the limit is 3/4...is that right?

 

[Dick]

That's right.