Calculating the Sum of an Infinite Series

In summary, the conversation is about evaluating the sum \sum\limits^\inf_{n=1} \frac{6}{n(n+1)} and the process of using partial fractions to solve it. The final part of the conversation is discussing how to find the formula for the nth term of the sum.
  • #1
theRukus
49
0

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.
 
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  • #2
theRukus said:

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
theRukus said:

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
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
theRukus said:
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
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
theRukus said:
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
theRukus said:
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
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
theRukus said:
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
Leave them grouped together and write out the first several terms and see if you notice anything.
 

1. How do you calculate the sum of an infinite series?

To calculate the sum of an infinite series, you need to use a specific formula called the summation formula. This formula involves finding the common ratio or difference between each term in the series, as well as the first term. Once you have these values, you can plug them into the formula and solve for the sum.

2. What is an infinite series?

An infinite series is a mathematical concept that involves adding an infinite number of terms together. These terms usually follow a specific pattern or rule, and the goal is to find the sum of all the terms in the series.

3. Can you use any formula to calculate the sum of an infinite series?

No, you cannot use any formula to calculate the sum of an infinite series. Each series may have its own unique formula or method for calculating the sum. It is important to identify the type of series and use the appropriate formula or method.

4. Are there different types of infinite series?

Yes, there are different types of infinite series, such as geometric series, arithmetic series, and power series. Each type of series has its own characteristics and methods for finding the sum.

5. What is the importance of calculating the sum of an infinite series?

Calculating the sum of an infinite series is important in various fields of mathematics, science, and engineering. It allows us to solve problems involving infinite quantities and continuous processes. It also helps in understanding the behavior and patterns of different series and their applications in real-world situations.

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