Telescoping Series involving 3 parts.

In summary, the problem involves determining the convergence or divergence of the series ##\displaystyle \sum_{n=2}^\infty \frac{1}{n^3-1}## by expressing it as a telescoping series. After partial fraction decomposition, the series is found to be equivalent to ##\displaystyle \sum_{n=2}^\infty \left(\frac{n}{n^2-1}-\frac{1}{n}\right)## and the cancellation pattern is found by writing out terms and combining them in pairs. The final sum is determined to be 1/4.
  • #1
B18
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0

Homework Statement


Determine whether the series is convergent or divergent by expressing s sub n as a telescoping series. If convergent find the sum.
[tex] \sum_{n=2}^\infty \frac{1}{n^3-1} [/tex]

The Attempt at a Solution


First thing I did was partial fraction decomposition.
Resulting in: Ʃ (n=2 to ∞) (-1/n+(1/2)/(n+1)+(1/2)/(n-1))
What I am confused with is when I try to find the partial sum nothing is canceling out. I tried to making it into this: Ʃ(n=2 to ∞) -1/n+(1/2)*Ʃ(n=2 to ∞) (1/n-1)+(1/n+1) but this didn't click with me either.
 
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  • #2
##\displaystyle \sum_{n=2}^\infty \frac{1}{n^3-n}##,

which you found to be equivalent to

##\displaystyle \sum_{n=2}^\infty \left(-\frac{1}{n}+\frac{1}{2}\left(\frac{1}{n+1}+\frac{1}{n-1}\right)\right)##

Or, equivalently,

##\displaystyle \sum_{n=2}^\infty \left(\frac{n}{n^2-1}-\frac{1}{n}\right)##

I'm not sure about the specifics of convergence/divergence for telescoping series, but maybe writing the series in this way will help.
 
  • #3
The part I am currently struggling with is when i find the partial sum and plug in some numbers nothing is canceling out. Still haven't figured this out yet...
 
  • #4
B18 said:
The part I am currently struggling with is when i find the partial sum and plug in some numbers nothing is canceling out. Still haven't figured this out yet...

n=2 gives -1/2+(1/6+1/2)
n=3 gives -1/3+(1/8+1/4)
n=4 gives -1/4+(1/10+1/6)
n=5 gives -1/5+(1/12+1/8)
...

Now notice that the 1/6 in the first line plus the 1/6 in the third line gives you 1/3 which cancels the -1/3 in the second line. Do you see the cancellation pattern now?
 
  • #5
So would the sum be :-1/2+1/2+1/4=1/4 If I am understanding the cancellation pattern correctly?
 
  • #6
B18 said:
So would the sum be :-1/2+1/2+1/4=1/4 If I am understanding the cancellation pattern correctly?

Yes, it should. Writing out a lot of terms and staring at it will usually show the pattern, plus the factor of (1/2) on some of them will suggest you need to combine some in pairs.
 
  • #7
Thank you sith and dick for your help I understand the problem now!
 

1. What is a telescoping series involving 3 parts?

A telescoping series involving 3 parts is a mathematical series in which terms cancel out in a pattern, leaving only a finite number of terms. It is called "telescoping" because the resulting sum can be simplified to just a few terms, similar to how a telescope collapses to a smaller size.

2. How do I identify if a series is a telescoping series involving 3 parts?

To identify if a series is a telescoping series involving 3 parts, you can look for a pattern in the terms where they cancel out. This can often be found by expanding the series and looking for common terms that can be simplified. Additionally, if the series involves fractions with terms that decrease in size, it is likely a telescoping series.

3. What is the formula for finding the sum of a telescoping series involving 3 parts?

The formula for finding the sum of a telescoping series involving 3 parts is S = a + b + c, where a, b, and c are the remaining terms after cancellation. This formula can also be written as S = a + (b - b1) + (c - c1), where b1 and c1 are the terms that canceled out in the series.

4. Can a telescoping series involving 3 parts have an infinite number of terms?

No, a telescoping series involving 3 parts cannot have an infinite number of terms because the cancellation pattern will eventually reach a finite number of terms. This is what makes it a telescoping series, as the number of terms "telescopes" or reduces to a smaller number.

5. Are there any real-life applications of telescoping series involving 3 parts?

Yes, telescoping series involving 3 parts can be used in various real-life applications, such as in physics and engineering to calculate the distance traveled or energy expended. They can also be used in finance and economics to calculate the present value of investments or annuities with a fixed rate of return.

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