Calculating the Sum of an Infinite Series

[theRukus]

Homework Statement

Evaluate the sum of the following:

$\sum\limits^\inf_{n=1} \frac{6}{n(n+1)}$

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.

 

[LCKurtz]

Homework Statement

Evaluate the sum of the following:

$\sum\limits^\inf_{n=1} \frac{6}{n(n+1)}$

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
$$\frac 6 {n(n+1)}= \frac A n + \frac B {n+1}$$
and write out the first n terms.

 

[Ray Vickson]

Homework Statement

Evaluate the sum of the following:

$\sum\limits^\inf_{n=1} \frac{6}{n(n+1)}$

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

 

[theRukus]

Alright, I've separated the equation into the following:

$\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$

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

So, I'm left with:

$\sum\limits^\infty_{n=1} \frac{6}{n}$

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!

 

[Dick]

Alright, I've separated the equation into the following:

$\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$

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

So, I'm left with:

$\sum\limits^\infty_{n=1} \frac{6}{n}$

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.

 

[theRukus]

Alright, well this is what I've done:

$\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$

$\frac{6}{n(n+1)} = \frac{A(n+1) + Bn}{n(n+1)}$

$6 = A(n+1) + Bn$

$6 = An + A + Bn$

Split into two equations,

$1: 6 = A + B$
$2: 6 = A$

Then,

$2->1: 6 = 6 + B$
$B = 0$

I can see I'm doing something wrong, because $\frac{6}{n}$ does not equal $\frac{6}{n(n+1)}$... What am I doing wrong?

 

[LCKurtz]

Alright, well this is what I've done:

$\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$

$\frac{6}{n(n+1)} = \frac{A(n+1) + Bn}{n(n+1)}$

$6 = A(n+1) + Bn$

$6 = An + A + Bn$

Split into two equations,

$1: 6 = A + B$
$2: 6 = A$

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?

 

[Dick]

Alright, well this is what I've done:

$\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$

$\frac{6}{n(n+1)} = \frac{A(n+1) + Bn}{n(n+1)}$

$6 = A(n+1) + Bn$

$6 = An + A + Bn$

Split into two equations,

$1: 6 = A + B$
$2: 6 = A$

Then,

$2->1: 6 = 6 + B$
$B = 0$

I can see I'm doing something wrong, because $\frac{6}{n}$ does not equal $\frac{6}{n(n+1)}$... 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.

 

[theRukus]

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

$\sum\limits^\infty_i=0 \frac{6}{n} - \frac{6}{n+1}$

$= \sum\limits^\infty_i \frac{6}{n} - \sum\limits^\infty_i \frac{6}{n+1}$

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..

 

[Ray Vickson]

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

$\sum\limits^\infty_i=0 \frac{6}{n} - \frac{6}{n+1}$

$= \sum\limits^\infty_i \frac{6}{n} - \sum\limits^\infty_i \frac{6}{n+1}$

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 $\sum_{n=1}^\infty [ \frac{6}{n} - \frac{6}{n+1} ],$ 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

 

[LCKurtz]

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