# Find the value approximately of 1/1*2+1/2*3+1/3*4

1. Jul 9, 2005

### stupidkid

find the value approximately of
1/1*2+1/2*3+1/3*4.........
Is there a simple way to find the anwer to problems like this?????

2. Jul 9, 2005

### TenaliRaman

If i understand your question correctly, then
1. do u know the sigma notation?
2. can you write your series using the sigma notation?
3. once you are done with step 2, can you do some simplification?
4. do u know what is a telescoping series?

-- AI

3. Jul 9, 2005

### philosophking

Oh man, I thought it was 1/(1*2)+1/(2*3) ... and that was really hard. That's so much easier though.

4. Jul 9, 2005

### Jameson

Hmmm... I guess it could be like that too. The OP will probably post if my interpretation was incorrect.

5. Jul 9, 2005

### Zurtex

Erm, that is really easy to work out and if it isn't that what else could it be, after all:

(1/1)*2 + (1/2)*3 + (1/3)*4 + ...

Does clearly not converge.

6. Jul 9, 2005

### Jameson

Doesn't converge? Would you explain why not? I see a convergence.

7. Jul 9, 2005

### LeonhardEuler

The terms aren't even approaching zero.

8. Jul 9, 2005

### Jameson

Right, but there is a cancellation.

9. Jul 9, 2005

### LeonhardEuler

What are you talking about? Every term is greater than 1, so the sum is greater than the number of terms. Therefore as the number of terms approaches infinity, so does the sum.

10. Jul 9, 2005

### Jameson

Woah, my bad guys. I totally looked at that the wrong way. Sorry about that. I'm tired.

The series should be.... $$a_{n}=\frac{1}{n*(n+1)}$$

Correct?

Last edited: Jul 9, 2005
11. Jul 9, 2005

### LeonhardEuler

Yes, I think thats what it is. He said he needed an approximation, so do you think he should just integrate it?

12. Jul 9, 2005

### mathwonk

just adding up a few terms makes it look really easy. but maybe i am too optimistic with small evidence.

13. Jul 9, 2005

### Curious3141

What is the need for approximation ? It's easy to get an exact answer.

1) Sigma notation
2) Partial fraction decomposition, splitting up the sum into the difference of two sums
3) Inspection and cancellation of a *lot* of terms (this is pertaining to Tenali's mention of telescoping series)
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$$