Finding the Sum of n/[(n+1)(n+2)(n+3)]

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Discussion Overview

The discussion revolves around finding the sum of the series Σn/[(n+1)(n+2)(n+3)]. Participants explore methods for evaluating this series, including partial fraction decomposition and telescoping series, while expressing varying levels of understanding and progress.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant requests assistance with the series Σn/[(n+1)(n+2)(n+3)].
  • Another suggests using partial fractions and mentions a method involving one function minus another.
  • A participant notes that while each fraction diverges, the overall sum converges, expressing uncertainty about the next steps.
  • There is a proposal to split the series into a form involving Σ1/[n+1] - 1/[n+2].
  • Some participants express confusion about the convergence to 1/4 and seek proof of this result.
  • Concerns are raised about the validity of taking certain terms outside the summation index.
  • One participant mentions the concept of a telescoping series and its relevance to the problem.
  • Another participant provides a method for determining constants A, B, and C in the partial fraction decomposition.
  • A later reply confirms the result of 1/4 and suggests that observing the pattern in the first few terms can lead to a general equation for the sum.

Areas of Agreement / Disagreement

Participants express differing views on the methods to solve the series, with some advocating for partial fractions and others questioning the manipulation of terms within the summation. There is no consensus on the best approach or the validity of certain steps.

Contextual Notes

Some participants indicate that the series may involve complexities related to convergence and the manipulation of summation indices, which remain unresolved.

Who May Find This Useful

Readers interested in series convergence, partial fraction decomposition, or telescoping series may find the discussion relevant.

Feynmanfan
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Could anybody please tell me how to do this?

Σn/[(n+1)(n+2)(n+3)]

Thanks!
 
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You tried splitting up into partial fractions and using the method where you have one function minus another?
 
I've already splitted it. It's interesting that each fraction alone diverges but the whole sum converges. But now I don't know what to do. What is this method you told me about?

Thanks
 
as zurtx said you can split it like this : Σ(1/[n+3])*(1/[n+1]-1/[n+2])=1/[n+3]*Σ1/[n+1]-1/[n+2]).
 
Still can't solve it. My computer says it certainly converges to 1/4. How can I prove that?
 
Feynmanfan said:
Still can't solve it. My computer says it certainly converges to 1/4. How can I prove that?

did you learned the term when a series is telescoped (or something of that that sort i have it on a book in hebrew in level of calclus 2)?
 
LQG, you can't take that term outside the summation as the n is part of the summation index.
 
matt grime said:
<acronym title='Loop Quantum Gravity' style='cursor:help;'>LGQ</acronym>, you can't take that term outside the summation as the n is part of the summation index.
it cannot be representing a coefficient in the form of Σd*(a1-an)
where d=1/(n+3) is the coefficient?
 
Apologies for getting LGQ and <acronym title='Loop Quantum Gravity' style='cursor:help;'>LGQ</acronym> mixed up (if it's any consolation I'm reading a paper on TQFT and such as I write (well, not exactly "as I write" but you get the idea))

You're taking the sum indexed by n when you say 1/[n+3]*Σ1/[n+1]-1/[n+2] ie n runs from something to something. You can't take the n+3 outside like that.

It's just like taking the x outside an integral wrt x (of course it's possible that it is the interior part of the summation that is wrong)
 
Last edited:
  • #10
matt grime is right. Yet that doesn't solve the problem. Thanks anyway for your help
 
  • #11
but you can do it by separating into partial fractions, it's quite easy isn't it?
 
  • #12
Where A, B, C are arbitrary constants let:

[tex]\frac{n}{(n+1)(n+2)(n+3)} = \frac{A}{n+1} + \frac{B}{n+2} + \frac{C}{n+3}[/tex]

Work out A, B and C. Now I could be wrong but I got A and C as negative numbers. So you should be able to write out the functions first few values n= 1, 2, 3, 4 and spot a pattern that if you add them together terms will start to cancel and you can extend that to n = r-3, r-2, r-1, r and get a general equation for the sum of f(n) from n = 1 to r.
 
  • #13
I just worked this out and the answer is in fact 1/4. Zurtex has the right approach. Find A, B and C and then write out the first several terms for each partial fraction. You should see a bunch of them cancel and you'll be left with a couple terms per fraction. From there, you can see the limit.
 
  • #14
Thanks everybody! You're all great.
 

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