What is the sum of the infinite series 1/[n(n+1)(n+2)]?

In summary, the infinite series defined by 1/[n(n+1)(n+2)] can be simplified by using partial fractions and rearranging the terms into a telescoping series. The partial sum can be found to be 1/4 by using Maple and applying the property of "psi". Another method of decomposition using logarithmic derivatives of the Gamma Euler function also leads to the same result.
  • #1
adamg
48
0
hey, i was just wondering if anyone could help find the sum of the infinite series defined by 1/[n(n+1)(n+2)]. I can split it into partial fractions but not sure from there. Thanks
 
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  • #2
its a/(1-r)...
where |r| < 1

where a is the initial values (1/6), and r is the common ratio...
it to late for my brain to think right now so I can tthink of what the ratio is going to be.. wil try tomorrow if nobody else has helped you.. going tp sleep now..gonight!
 
  • #3
I'm not sure the splitting can do you any good,because each of the series you'd be geting would be infinite.If you don't know the answer,well,my Maple said it's 1/4...i don't know how it did it...

I'm going to check it in G & R,too...

Daniel.
 
  • #4
i thought you could do it by splitting it into partial fractions, then finding an expression for the nth partial sum and let n tend to infinity
 
  • #5
According to Maple,the partial sum is:
[tex] \sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)} =-\frac{1}{2(n+2)(n+1)}+\frac{1}{4} [/tex]

If you can find a way to prove what I've just written,then that's it...

Daniel.
 
  • #6
I think partial fractions is promising too -- with care, you can probably arrange the terms into one or more telescoping series. What did you get when you applied partial fractions?
 
  • #7
Here's the proof for the partial sum:

[tex] S=:\sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)}=\frac{1}{4}-\frac{1}{2(n+1)(n+2)} [/tex](1)
-----------------------------||------------------------

Use partial fractions to rewrite the initial partial sum as:

[tex] S=\sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)}=\sum_{k=1}^{n}\frac{1}{2k}-\sum_{k=1}^{n}\frac{1}{k+1}+\sum_{k=1}^{n}\frac{1}{2(k+2)} [/tex] (2)

According to Maple each of the three sums is:

[tex] \sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\psi(n+1)+\frac{1}{2}\gamma [/tex] (3)

[tex] \sum_{k=1}^{n} \frac{-1}{k+1}=-\psi(n+2)+1-\gamma [/tex] (4)

[tex] \sum_{k=2}^{n} \frac{1}{2(k+2)}=\frac{1}{2}\psi(n+3)-\frac{3}{4}+\frac{1}{2}\gamma [/tex](5)

Add (3)->(5) and equate with (2):

[tex] S=\frac {1}{4}+\frac{1}{2}\psi(n+1)-\psi(n+2)+\frac{1}{2}\psi(n+3) [/tex](6)

Now,use the property of "psi":

[tex] \psi(z+1)=\psi(z)+\frac{1}{z} [/tex] (7) for [itex]z\neq 0 [/itex]

to write:

[tex] \psi(n+2)=\psi(n+1)+\frac{1}{n+1} [/tex] (8)

[tex] \psi(n+3)=\psi(n+2)+\frac{1}{n+2}=\psi(n+1)+\frac{1}{n+1}+\frac{1}{n+2} [/tex] (9)

to rewrite the "3-psi" term from (6) simply as:

[tex] -\frac{1}{2(n+1)(n+2)} [/tex] (10)

Then add (10) to 1/4 from (6) to get the equality you were supposed to prove... :wink:

In the initial equality set [itex] n\rightarrow +\infty [/itex] to get the answer [itex] \frac{1}{4} [/itex]

Daniel.
 
  • #8
What do you think of this decomposition?

[tex]
\frac{1}{k (k+1) (k+2)} = \frac{1}{2}
\left(
\frac{1}{k} - \frac{1}{k+1} - \frac{1}{k+1} + \frac{1}{k+2}
\right)
[/tex]
 
  • #9
That one gives the answer,too...1/4...

Daniel.
 
  • #10
Hurkyl said:
What do you think of this decomposition?

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


i got this far, then not sure how to cancel terms to find the sum?
 
  • #11
Well,it's not difficult.You may write it as
[tex] \frac{1}{2}[(\frac{1}{k}-\frac{1}{k+1})-(\frac{1}{k+1}-\frac{1}{k+2})] [/tex]

and then give values to "k" to see the pattern the terms take...

Daniel.
 
  • #12
yeah got it now, thanks for all the help!
 
  • #13
dextercioby said:
According to Maple each of the three sums is:

[tex] \sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\psi(n+1)+\frac{1}{2}\gamma [/tex] (3)

[tex] \sum_{k=1}^{n} \frac{-1}{k+1}=-\psi(n+2)+1-\gamma [/tex] (4)

[tex] \sum_{k=2}^{n} \frac{1}{2(k+2)}=\frac{1}{2}\psi(n+3)-\frac{3}{4}+\frac{1}{2}\gamma [/tex](5)

Daniel.

Thanks, didn't have a clue. Turns out the first one is the following:

[tex]\sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\{\frac{\int_0^\infty e^{-t}t^{n}\ln(t)dt}{\int_0^\infty e^{-t}t^{n}dt}+\int_0^\infty \frac{1}{t}(\frac{1}{1+t}-e^{-t})dt\} [/tex]

I had no idea! Does Adamg know this? Don't mind me, I mean I had problems with Leibnitz's rule the other day. I tell you what though, if I was the teacher assigning this problem, we'd be looking at these integrals. Now how in the world does the sum turn out to be this messy expression?
 
  • #14
It comes from the logarithmic derivative of the Gamma Euler function,which is "psi"...

Daniel.
 
  • #15
Is an easy sum:

[tex]\sum\frac{1}{n(n+1)(n+2)}[/tex]=[tex]\sum\frac{1}{2n}[/tex]+[tex]\frac{1}{2(n+2)}[/tex]-[tex]\frac{1}{n+1}[/tex]= ([tex]\frac{1}{2}[/tex]+[tex]\frac{1}{4}[/tex]+[tex]\frac{1}{6}[/tex]+[tex]\frac{1}{8}[/tex]+[tex]\frac{1}{10}[/tex]...+[tex]\frac{1}{6}[/tex]+[tex]\frac{1}{8}[/tex]+[tex]\frac{1}{10}[/tex]+...+(-[tex]\frac{1}{2}[/tex]-[tex]\frac{1}{3}[/tex]-[tex]\frac{1}{4}[/tex]-[tex]\frac{1}{5}[/tex]-[tex]\frac{1}{6}[/tex]-...) = [tex]\frac{1}{2}[/tex]+[tex]\frac{1}{4}[/tex]+[tex]\frac{1}{3}[/tex]+[tex]\frac{1}{4}[/tex]+[tex]\frac{1}{5}[/tex]+... +(-[tex]\frac{1}{2}[/tex]-[tex]\frac{1}{3}[/tex]-[tex]\frac{1}{4}[/tex]-[tex]\frac{1}{5}[/tex]-[tex]\frac{1}{6}[/tex]-...) = [tex]\frac{1}{4}[/tex].
 

1. What is the sum of an infinite series?

The sum of an infinite series is the total value obtained by adding all the terms in the series together, from the first term to infinity.

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

To calculate the sum of an infinite series, you can use various methods such as the geometric series formula or the telescoping series method. These methods involve finding a pattern in the series and using mathematical formulas to determine the sum.

3. Can the sum of an infinite series be negative?

Yes, the sum of an infinite series can be negative. This can happen when the terms in the series alternate between positive and negative values, resulting in a negative overall sum.

4. What is the significance of the sum of an infinite series?

The sum of an infinite series is important in mathematics as it helps us understand the behavior of a series and determine if it converges or diverges. It also has applications in various fields such as physics, engineering, and economics.

5. Can the sum of an infinite series be infinite?

Yes, the sum of an infinite series can be infinite. This happens when the terms in the series increase in value without bound, resulting in a sum that also increases without bound. In this case, the series is said to diverge to infinity.

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