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

  • Thread starter stupidkid
  • Start date
  • #1
stupidkid
18
0
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?????
 

Answers and Replies

  • #2
TenaliRaman
644
1
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
philosophking
175
0
Oh man, I thought it was 1/(1*2)+1/(2*3) ... and that was really hard. That's so much easier though.
 
  • #4
Jameson
Gold Member
MHB
4,468
4
Hmmm... I guess it could be like that too. The OP will probably post if my interpretation was incorrect.
 
  • #5
Zurtex
Science Advisor
Homework Helper
1,120
1
philosophking said:
Oh man, I thought it was 1/(1*2)+1/(2*3) ... and that was really hard. That's so much easier though.
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
Jameson
Gold Member
MHB
4,468
4
Doesn't converge? Would you explain why not? I see a convergence.
 
  • #7
LeonhardEuler
Gold Member
859
1
The terms aren't even approaching zero.
 
  • #8
Jameson
Gold Member
MHB
4,468
4
Right, but there is a cancellation.
 
  • #9
LeonhardEuler
Gold Member
859
1
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
Jameson
Gold Member
MHB
4,468
4
Woah, my bad guys. I totally looked at that the wrong way. Sorry about that. I'm tired.

The series should be.... [tex]a_{n}=\frac{1}{n*(n+1)}[/tex]

Correct?
 
Last edited by a moderator:
  • #11
LeonhardEuler
Gold Member
859
1
Yes, I think thats what it is. He said he needed an approximation, so do you think he should just integrate it?
 
  • #12
mathwonk
Science Advisor
Homework Helper
11,352
1,584
just adding up a few terms makes it look really easy. but maybe i am too optimistic with small evidence.
 
  • #13
Curious3141
Homework Helper
2,850
88
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)
4) The answer, the end.
 
Last edited:
  • #14
Zurtex
Science Advisor
Homework Helper
1,120
1
mathwonk said:
just adding up a few terms makes it look really easy. but maybe i am too optimistic with small evidence.
If the sum is:

[tex]\sum_{n=1}^{\infty} \frac{1}{n(n+1)}[/tex]

Then the solution is really simple. But the poster hasn't posted what they know, so it's not like we can do it for them....
 

Suggested for: Find the value approximately of 1/1*2+1/2*3+1/3*4

Replies
1
Views
804
Replies
1
Views
762
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
3K
Replies
6
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
483
  • Last Post
Replies
4
Views
2K
Top