Telescoping Series

[nasi112]

I have solved tons of telescoping series, and it was always easy to recognize the cancellation pattern, but in this one I don't know why it is not obvious.


$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k(k+1)(k+2)} = \sum_{k=1}^{\infty}\frac{1}{2k} + \frac{1}{k + 2} - \frac{1}{2k + 4} - \frac{1}{k + 1}$

$\displaystyle =\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{1}{2k} + \frac{1}{k + 2} - \frac{1}{2k + 4} - \frac{1}{k + 1}$

$\displaystyle =\lim_{n\rightarrow \infty}\left(\left[\frac{1}{2} + \frac{1}{3} - \frac{1}{6} - \frac{1}{2}+\right] + \left[\frac{1}{4} + \frac{1}{4} - \frac{1}{8} - \frac{1}{3}\right] + ...... + \frac{1}{2n} + \frac{1}{n + 2} - \frac{1}{2n + 4} - \frac{1}{n + 1}\right)$

Magically, everything will get cancelled except one term:

$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k(k+1)(k+2)} =\frac{1}{4}$

Should I keep adding more terms to see the cancellation pattern? If you have a better way to solve this series, sharpen your pencil, and show us how you will do it.

 

[Steven G]

Your very first equal sign is wrong.

 

[Dr.Peterson]

I have solved tons of telescoping series, and it was always easy to recognize the cancellation pattern, but in this one I don't know why it is not obvious.


$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k(k+1)(k+2)} = \sum_{k=1}^{\infty}\frac{1}{2k} + \frac{1}{k + 2} - \frac{1}{2k + 4} - \frac{1}{k + 1}$

$\displaystyle =\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{1}{2k} + \frac{1}{k + 2} - \frac{1}{2k + 4} - \frac{1}{k + 1}$

$\displaystyle =\lim_{n\rightarrow \infty}\left(\left[\frac{1}{2} + \frac{1}{3} - \frac{1}{6} - \frac{1}{2}+\right] + \left[\frac{1}{4} + \frac{1}{4} - \frac{1}{8} - \frac{1}{3}\right] + ...... + \frac{1}{2n} + \frac{1}{n + 2} - \frac{1}{2n + 4} - \frac{1}{n + 1}\right)$

Magically, everything will get cancelled except one term:

$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k(k+1)(k+2)} =\frac{1}{4}$

Should I keep adding more terms to see the cancellation pattern? If you have a better way to solve this series, sharpen your pencil, and show us how you will do it.

Here's how I worked it out (trusting your first step):

The denominator (+) 2k will be 2, 4, 6, 8, 10, ...
The denominator (-) 2k+4 will be 6, 8, 10, ...
Only terms with 2 and 4 are left uncancelled.

The denominator (+) k+2 will be 3, 4, 5, 6, 7, 8, 9, 10, ...
The denominator (-) k+1 will be 2, 3, 4, 5, 6, 7, 8, 9, 10, ...
Everything is cancelled, and the 2 cancels the 2 that hadn't been cancelled above.

So only the term with denominator 4 remains.

 

[Steven G]

How can partial decomposition with a denominator with three factors become a sum of four fractions????

 

[skeeter]

$\displaystyle \sum_{k=1}^\infty \dfrac{1}{k(k+1)(k+2)} = \dfrac{1}{2} \bigg[\sum_{k=1}^\infty \dfrac{1}{k}-\dfrac{1}{k+1} - \sum_{k=1}^\infty \dfrac{1}{k+1}-\dfrac{1}{k+2} \bigg]$

 

[Dr.Peterson]

I have solved tons of telescoping series, and it was always easy to recognize the cancellation pattern, but in this one I don't know why it is not obvious.


$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k(k+1)(k+2)} = \sum_{k=1}^{\infty}\frac{1}{2k} + \frac{1}{k + 2} - \frac{1}{2k + 4} - \frac{1}{k + 1}$

Since it's been pointed out that your first line is unconventional, it's worth asking how you got that, rather than the usual three terms. A different first step might have helped you do the rest more easily. Was that step given to you?

@skeeter's approach nicely takes the three terms you would normally get, and rearranges them into a pair of naturally telescoping series.

Magically, everything will get cancelled except one term:

I also need to ask, did you yourself "magically" get the 1/4, or were you saying someone else told you that it should, and you don't see how? I prefer not to say something is true until I have reason to be sure (without magic).

 

[blamocur]

How can partial decomposition with a denominator with three factors become a sum of four fractions????

Two of them have similar first order denominators, i.e., k+1 and 2k+4. But I agree that reducing it to a sum of three fractions makes it more manageable.

 

[blamocur]

I have solved tons of telescoping series, and it was always easy to recognize the cancellation pattern, but in this one I don't know why it is not obvious.


$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k(k+1)(k+2)} = \sum_{k=1}^{\infty}\frac{1}{2k} + \frac{1}{k + 2} - \frac{1}{2k + 4} - \frac{1}{k + 1}$

$\displaystyle =\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{1}{2k} + \frac{1}{k + 2} - \frac{1}{2k + 4} - \frac{1}{k + 1}$

$\displaystyle =\lim_{n\rightarrow \infty}\left(\left[\frac{1}{2} + \frac{1}{3} - \frac{1}{6} - \frac{1}{2}+\right] + \left[\frac{1}{4} + \frac{1}{4} - \frac{1}{8} - \frac{1}{3}\right] + ...... + \frac{1}{2n} + \frac{1}{n + 2} - \frac{1}{2n + 4} - \frac{1}{n + 1}\right)$

Magically, everything will get cancelled except one term:

$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k(k+1)(k+2)} =\frac{1}{4}$

Should I keep adding more terms to see the cancellation pattern? If you have a better way to solve this series, sharpen your pencil, and show us how you will do it.

To break the original term into a sum of three fractions you can simply write an equation:
$$\frac{1}{k(k+1)(k+2)} = \frac{x}{k} + \frac{y}{k+1} + \frac{z}{k+2}$$and solve it for $x,y,z$. You will actually get a sum which is equivalent to the one you've already gotten, but it will look more manageable.

 

[skeeter]

$\dfrac{1}{k(k+1)(k+2)} = \dfrac{A}{k} + \dfrac{B}{k+1} + \dfrac{C}{k+2}$

$1 = A(k+1)(k+2) + Bk(k+2) + Ck(k+1)$

$k=0 \implies 1=2A \implies A=\dfrac{1}{2}$

$k=-1 \implies 1 = -B \implies B =-1$

$k=-2 \implies 1= 2C \implies C = \dfrac{1}{2}$

$\dfrac{1}{2k} - \dfrac{1}{k+1} + \dfrac{1}{2(k+2)} = \dfrac{1}{2} \bigg[\dfrac{1}{k} - \dfrac{2}{k+1} + \dfrac{1}{k+2} \bigg] = \dfrac{1}{2} \bigg[\left(\dfrac{1}{k} - \dfrac{1}{k+ 1}\right) - \left(\dfrac{1}{k+1} - \dfrac{1}{k+2} \right) \bigg]$

 

[nasi112]

Thanks a lot to everyone who entered in this post and gave something useful.

Your very first equal sign is wrong.

Where exactly?

Here's how I worked it out (trusting your first step):

The denominator (+) 2k will be 2, 4, 6, 8, 10, ...
The denominator (-) 2k+4 will be 6, 8, 10, ...
Only terms with 2 and 4 are left uncancelled.

The denominator (+) k+2 will be 3, 4, 5, 6, 7, 8, 9, 10, ...
The denominator (-) k+1 will be 2, 3, 4, 5, 6, 7, 8, 9, 10, ...
Everything is cancelled, and the 2 cancels the 2 that hadn't been cancelled above.

So only the term with denominator 4 remains.

This is a very unique way. It is a big shortcut to cancel terms easily. Nice solution Dr. Peterson.

How can partial decomposition with a denominator with three factors become a sum of four fractions????

I prefer to do partial fraction with only two fractions at once. For example:

$\displaystyle \frac{1}{k(k+1)(k+2)} = \frac{1}{(k + 2)}\left[\frac{1}{k(k + 1)}\right]$

First I work on this.

$\displaystyle \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1}$

$\displaystyle \frac{1}{(k + 2)}\left[\frac{1}{k(k + 1)}\right] = \frac{1}{k(k + 2)} - \frac{1}{(k + 1)(k + 2)}$

Then, I repeat the process again for each separately. This is why you see 4 terms.

$\displaystyle \sum_{k=1}^\infty \dfrac{1}{k(k+1)(k+2)} = \dfrac{1}{2} \bigg[\sum_{k=1}^\infty \dfrac{1}{k}-\dfrac{1}{k+1} - \sum_{k=1}^\infty \dfrac{1}{k+1}-\dfrac{1}{k+2} \bigg]$

This is similar to what I have done. Nice job skeeter.

Since it's been pointed out that your first line is unconventional, it's worth asking how you got that, rather than the usual three terms. A different first step might have helped you do the rest more easily. Was that step given to you?

@skeeter's approach nicely takes the three terms you would normally get, and rearranges them into a pair of naturally telescoping series.

I also need to ask, did you yourself "magically" get the 1/4, or were you saying someone else told you that it should, and you don't see how? I prefer not to say something is true until I have reason to be sure (without magic).

I hope that my explanation to Professor Steven has answered your first question. For the second question, I have tested the telescoping series in wolfram and got the 1/4 magically there, but with your explanation got it why all terms got cancelled except this 1/4.

To break the original term into a sum of three fractions you can simply write an equation:
$$\frac{1}{k(k+1)(k+2)} = \frac{x}{k} + \frac{y}{k+1} + \frac{z}{k+2}$$and solve it for $x,y,z$. You will actually get a sum which is equivalent to the one you've already gotten, but it will look more manageable.

I should have done it this way. In fact this is faster, but I used to do each two fractions at once. I have never thought my way would make genius professors like you get confused.

$\dfrac{1}{k(k+1)(k+2)} = \dfrac{A}{k} + \dfrac{B}{k+1} + \dfrac{C}{k+2}$

$1 = A(k+1)(k+2) + Bk(k+2) + Ck(k+1)$

$k=0 \implies 1=2A \implies A=\dfrac{1}{2}$

$k=-1 \implies 1 = -B \implies B =-1$

$k=-2 \implies 1= 2C \implies C = \dfrac{1}{2}$

$\dfrac{1}{2k} - \dfrac{1}{k+1} + \dfrac{1}{2(k+2)} = \dfrac{1}{2} \bigg[\dfrac{1}{k} - \dfrac{2}{k+1} + \dfrac{1}{k+2} \bigg] = \dfrac{1}{2} \bigg[\left(\dfrac{1}{k} - \dfrac{1}{k+ 1}\right) - \left(\dfrac{1}{k+1} - \dfrac{1}{k+2} \right) \bigg]$

Nice steps skeeter. It seems that I will change my habit of solving problems the hard way.

 

[Steven G]

Not sure why my algebra did not show what you wrote is correct, but it is. Sorry!