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Sum of Convergent Series Help Please!

  • Thread starter BuBbLeS01
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  • #1
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1. Homework Statement
Find the sum of the convergent series:
The sum of 1/ (n^2 - 1) from n=2 to infinity


2. Homework Equations



3. 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...
 

Answers and Replies

  • #2
HallsofIvy
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1. Homework Statement
Find the sum of the convergent series:
The sum of 1/ (n^2 - 1) from n=2 to infinity


2. Homework Equations



3. 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
[tex]\frac{1}{n^2-1}= \frac{A}{n-1}+ \frac{B}{n+1}[/tex]
what are A and B?
 
  • #3
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A would be 1 * (n+1) = n+1
B would be 1 * (n-1) = n-1
right? or no??
 
  • #4
Dick
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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?
 
  • #5
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So I would have
A(n+1) * B(n-1)
 
  • #6
Dick
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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.
 
  • #7
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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)
 
  • #8
Dick
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You are getting closer, but you didn't multiply the left side by (n^2-1), did you?
 
  • #9
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Oh I didn't realize I had to....
So I get...
1 = A(n+1) + B(n-1)
 
  • #10
Dick
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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?
 
  • #11
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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???
 
  • #12
Dick
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Sure. Plugging in values of n will work too. But plugging in just n=1 doesn't tell you what B is, does it?
 
  • #13
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can't he plug in -1 for to zero out A and get B

(heaviside cover up method?)
 
  • #14
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Yea thats what I meant 1 to get A and -1 to get B
 
  • #15
Dick
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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?
 
  • #16
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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...
 
  • #17
Dick
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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?
 
  • #18
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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???
 
  • #19
Dick
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I'm not sure why you are saying 'first' and 'last' should cancel, but you are right. .5/2 doesn't cancel.
 
  • #20
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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.
 
  • #21
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so as n approaches infinity the limit is 3/4...is that right?
 
  • #22
Dick
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That's right.
 

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