# Series (help!)

1. Apr 7, 2004

### Feynmanfan

Could anybody please tell me how to do this?

Σn/[(n+1)(n+2)(n+3)]

Thanks!

2. Apr 7, 2004

### Zurtex

You tried splitting up into partial fractions and using the method where you have one function minus another?

3. Apr 7, 2004

### Feynmanfan

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. Apr 7, 2004

### MathematicalPhysicist

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. Apr 7, 2004

### Feynmanfan

Still can't solve it. My computer says it certainly converges to 1/4. How can I prove that?

6. Apr 7, 2004

### MathematicalPhysicist

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. Apr 7, 2004

### matt grime

LQG, you can't take that term outside the summation as the n is part of the summation index.

8. Apr 7, 2004

### MathematicalPhysicist

it cannot be representing a coefficient in the form of Σd*(a1-an)
where d=1/(n+3) is the coefficient?

9. Apr 7, 2004

### matt grime

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: Apr 7, 2004
10. Apr 7, 2004

### Feynmanfan

matt grime is right. Yet that doesn't solve the problem. Thanks anyway for your help

11. Apr 7, 2004

### matt grime

but you can do it by separating into partial fractions, it's quite easy isn't it?

12. Apr 7, 2004

### Zurtex

Where A, B, C are arbitrary constants let:

$$\frac{n}{(n+1)(n+2)(n+3)} = \frac{A}{n+1} + \frac{B}{n+2} + \frac{C}{n+3}$$

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. Apr 8, 2004

### Anteros

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. Apr 9, 2004

### Feynmanfan

Thanks everybody! You're all great.