Understanding Telescoping Series: Finding the Sum

In summary, the poster asked for help finding the sum of an infinite series involving fractions. They approached it using a telescoping series and partial fraction decomposition, but could not find a discernible pattern. Another user suggested looking for a cancellation pattern, which the poster eventually found by stretching out the nth term values.
  • #1
qeteshchl
2
0
My apologies beforehand for not using the right format for this post.

Homework Statement



Find the sum of (from 1 to inf) of [tex]\sum[/tex]8/(n(n+1)(n+2))

Homework Equations





The Attempt at a Solution



I approached the problem like I would a telescoping series by using partial faction decomposition to split it up. I arrived at:

[tex]\sum[/tex]4/n - 8/(n+1) + 4/(n+2)

I started plugging in numbers for 1 to try and arrive at a pattern:

(1 - 1 + 1/3) + (1/2 - 2/3 + 1/4) + (1/3 - 2/4 + 1/5) + (1/4 - 2/5 + 1/6) ...

I'm not seeing any discernible pattern to what is canceling out. Am I just approaching this problem wrong? Thanks for any help you guys may provide!
 
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  • #2
qeteshchl said:
My apologies beforehand for not using the right format for this post.

Homework Statement



Find the sum of (from 1 to inf) of [tex]\sum[/tex]8/(n(n+1)(n+2))

Homework Equations


The Attempt at a Solution



I approached the problem like I would a telescoping series by using partial faction decomposition to split it up. I arrived at:

[tex]\sum[/tex]4/n - 8/(n+1) + 4/(n+2)

I started plugging in numbers for 1 to try and arrive at a pattern:

(1 - 1 + 1/3) + (1/2 - 2/3 + 1/4) + (1/3 - 2/4 + 1/5) + (1/4 - 2/5 + 1/6) ...

I'm not seeing any discernible pattern to what is canceling out. Am I just approaching this problem wrong? Thanks for any help you guys may provide!

I'm assuming your partial fraction expansion is OK. What I notice in your general term is the two 4's and the -8. If they went with the same n, they would cancel. But the 4/n will be 4/(n+1) in the next term and the 4/(n+2) will be 4/(n+1) in the previous term. Try looking for that cancellation pattern.
 
Last edited:
  • #3
I don't know if it was the most efficient way of doing it but I stretched out the nth term values pretty far until I found a pattern. Thanks for your advice LCKurtz, really appreciate the response!
 

FAQ: Understanding Telescoping Series: Finding the Sum

What is a telescoping series?

A telescoping series is a type of infinite series where each term is canceled out by the following term, leaving only a finite number of terms.

How do you find the sum of a telescoping series?

To find the sum of a telescoping series, you can use the formula S = lim n → ∞ Sn, where Sn is the partial sum of the first n terms of the series. If the limit exists, it will give you the sum of the infinite series.

What is the telescoping property of a series?

The telescoping property of a series refers to the fact that each term in the series can be cancelled out by the following term, leaving only a finite number of terms in the sum.

Can all series be telescoping series?

No, not all series can be telescoping series. A telescoping series must have a specific form where each term can be cancelled out, which is not always the case for other types of series.

What are some real-life applications of telescoping series?

Telescoping series have many practical applications in fields such as physics, engineering, and economics. For example, they can be used to calculate the distance traveled by an object with varying speeds, the total cost of a loan with changing interest rates, or the displacement of a vibrating object over time.

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