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Sum of Series

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  • #1
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Homework Statement


Evaluate the sum of the following:

[itex]\sum\limits^\inf_{n=1} \frac{6}{n(n+1)}[/itex]


Homework Equations





The Attempt at a Solution


Well... The denominator is going to get infinitely large as n approaches infinity, so would the value of the sum not converge to zero? The homework program said this answer was wrong.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Evaluate the sum of the following:

[itex]\sum\limits^\inf_{n=1} \frac{6}{n(n+1)}[/itex]


Homework Equations





The Attempt at a Solution


Well... The denominator is going to get infinitely large as n approaches infinity, so would the value of the sum not converge to zero? The homework program said this answer was wrong.
The first term is 3 and they are all positive, so how could the sum be zero?

Hint: Use partial fractions
[tex]\frac 6 {n(n+1)}= \frac A n + \frac B {n+1}[/tex]
and write out the first n terms.
 
  • #3
Ray Vickson
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Homework Statement


Evaluate the sum of the following:

[itex]\sum\limits^\inf_{n=1} \frac{6}{n(n+1)}[/itex]


Homework Equations





The Attempt at a Solution


Well... The denominator is going to get infinitely large as n approaches infinity, so would the value of the sum not converge to zero? The homework program said this answer was wrong.
So, by your reasoning the sum S = 1/2 + 1/4 + 1/8 + 1/16 + ... = Ʃ 1/2n is zero, because the terms are going to zero. Does that look right to you?

RGV
 
  • #4
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Alright, I've separated the equation into the following:

[itex]\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}[/itex]

And solved for A, B, getting A = 6, B = 0.

So, I'm left with:

[itex]\sum\limits^\infty_{n=1} \frac{6}{n}[/itex]

Now, I'm lost as to what to do. I don't know how to solve for this sum...

Any hints?
Thank you so much PhysicsForums!
 
  • #5
Dick
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Alright, I've separated the equation into the following:

[itex]\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}[/itex]

And solved for A, B, getting A = 6, B = 0.

So, I'm left with:

[itex]\sum\limits^\infty_{n=1} \frac{6}{n}[/itex]

Now, I'm lost as to what to do. I don't know how to solve for this sum...

Any hints?
Thank you so much PhysicsForums!
Your partial fraction calculation is obviously wrong. 6/n isn't equal to 6/(n*(n+1)). Try and notice things like that! Try it again. Show your work if you can't get it right.
 
  • #6
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Alright, well this is what I've done:

[itex]\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}[/itex]

[itex]\frac{6}{n(n+1)} = \frac{A(n+1) + Bn}{n(n+1)}[/itex]

[itex]6 = A(n+1) + Bn[/itex]

[itex]6 = An + A + Bn[/itex]

Split into two equations,

[itex]1: 6 = A + B[/itex]
[itex]2: 6 = A[/itex]

Then,

[itex]2->1: 6 = 6 + B[/itex]
[itex]B = 0[/itex]

I can see I'm doing something wrong, because [itex]\frac{6}{n}[/itex] does not equal [itex]\frac{6}{n(n+1)}[/itex].... What am I doing wrong?
 
  • #7
LCKurtz
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Alright, well this is what I've done:

[itex]\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}[/itex]

[itex]\frac{6}{n(n+1)} = \frac{A(n+1) + Bn}{n(n+1)}[/itex]

[itex]6 = A(n+1) + Bn[/itex]

[itex]6 = An + A + Bn[/itex]

Split into two equations,

[itex]1: 6 = A + B[/itex]
[itex]2: 6 = A[/itex]
Before you split it into two equations write the 3rd equation above like this:

0n + 6 = (A+B)n + A

So what should you get for A+B by equating coefficients?
 
  • #8
Dick
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Alright, well this is what I've done:

[itex]\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}[/itex]

[itex]\frac{6}{n(n+1)} = \frac{A(n+1) + Bn}{n(n+1)}[/itex]

[itex]6 = A(n+1) + Bn[/itex]

[itex]6 = An + A + Bn[/itex]

Split into two equations,

[itex]1: 6 = A + B[/itex]
[itex]2: 6 = A[/itex]

Then,

[itex]2->1: 6 = 6 + B[/itex]
[itex]B = 0[/itex]

I can see I'm doing something wrong, because [itex]\frac{6}{n}[/itex] does not equal [itex]\frac{6}{n(n+1)}[/itex].... What am I doing wrong?
Your first equation should be 0=A+B not 6=A+B. The n's need to cancel if you are always going to get 6.
 
  • #9
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Ah, perfect. I love you guys, you're so helpful. I guess I need to be a little more critical of my own work though, before I give up on what I've done completely.

Now though, knowing A & B (I guess I was solving for them wrong), I am still confused as to how to get the 'nth term' of a series.

So I found that A=6, B=-6. Then, I'm looking for

[itex]\sum\limits^\infty_i=0 \frac{6}{n} - \frac{6}{n+1}[/itex]

[itex] = \sum\limits^\infty_i \frac{6}{n} - \sum\limits^\infty_i \frac{6}{n+1}[/itex]

I don't know how to get the formula for the nth term of these sums, but I do know how to solve once I get that formula. Can someone give me a push?

Don't worry, I'm not just waiting for help from you guys.. I'm searching for tutorials on the Internet as well.. I missed my class on series / sums and don't know what I'm doing ... Thank you so much..
 
  • #10
Ray Vickson
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Ah, perfect. I love you guys, you're so helpful. I guess I need to be a little more critical of my own work though, before I give up on what I've done completely.

Now though, knowing A & B (I guess I was solving for them wrong), I am still confused as to how to get the 'nth term' of a series.

So I found that A=6, B=-6. Then, I'm looking for

[itex]\sum\limits^\infty_i=0 \frac{6}{n} - \frac{6}{n+1}[/itex]

[itex] = \sum\limits^\infty_i \frac{6}{n} - \sum\limits^\infty_i \frac{6}{n+1}[/itex]

I don't know how to get the formula for the nth term of these sums, but I do know how to solve once I get that formula. Can someone give me a push?

Don't worry, I'm not just waiting for help from you guys.. I'm searching for tutorials on the Internet as well.. I missed my class on series / sums and don't know what I'm doing ... Thank you so much..
You should have [itex] \sum_{n=1}^\infty [ \frac{6}{n} - \frac{6}{n+1} ], [/itex] not what you wrote; go back and look carefully at what you wrote. Now just write down the sum for n going from 1 to N for a few small values of N, to see what is happening.

RGV
 
  • #11
LCKurtz
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Leave them grouped together and write out the first several terms and see if you notice anything.
 

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