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

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In summary, zurtx's method works for solving a set of partial fractions where each fraction is divided by the coefficient d. You can also use the sum indexed by n, but zurtx's method is easier.
  • #1
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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  • #2
You tried splitting up into partial fractions and using the method where you have one function minus another?
 
  • #3
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
 
  • #4
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]).
 
  • #5
Still can't solve it. My computer says it certainly converges to 1/4. How can I prove that?
 
  • #6
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)?
 
  • #7
LQG, you can't take that term outside the summation as the n is part of the summation index.
 
  • #8
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?
 
  • #9
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)
 
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  • #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.
 

1. How do you find the sum of n/[(n+1)(n+2)(n+3)]?

To find the sum of n/[(n+1)(n+2)(n+3)], we can use the telescoping series method. We start by writing out the first few terms of the series, and then we use partial fraction decomposition to simplify the expression. After that, we take the limit as n approaches infinity to find the sum.

2. What is the telescoping series method?

The telescoping series method is a technique used to simplify infinite series by eliminating most of the terms. This is achieved by writing out the first few terms of the series and then using partial fraction decomposition to simplify the expression. The remaining terms will then "cancel out" or "collapse" into a simpler expression, making it easier to find the sum.

3. What is partial fraction decomposition?

Partial fraction decomposition is a method used to simplify rational expressions by breaking them down into simpler fractions. This is done by writing the expression as a sum of simpler fractions with distinct denominators. This technique is helpful in simplifying expressions and solving for unknown variables.

4. Why do we take the limit as n approaches infinity?

We take the limit as n approaches infinity to find the sum of the series because we are dealing with an infinite number of terms. This means that we need to find the value of the series as n gets larger and larger, approaching infinity. By taking the limit, we are essentially finding the "end behavior" of the series, which gives us the sum.

5. Can the sum of n/[(n+1)(n+2)(n+3)] be found without using the telescoping series method?

Yes, the sum can also be found by using other methods such as the geometric series method or the ratio test. However, the telescoping series method is often the most efficient and straightforward way to find the sum of this particular series. It is also a commonly used technique in mathematics and is worth understanding and practicing.

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