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

  • Thread starter stupidkid
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  • #1
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Main Question or Discussion Point

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
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
Oh man, I thought it was 1/(1*2)+1/(2*3) ... and that was really hard. That's so much easier though.
 
  • #4
789
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Hmmm... I guess it could be like that too. The OP will probably post if my interpretation was incorrect.
 
  • #5
Zurtex
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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
789
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Doesn't converge? Would you explain why not? I see a convergence.
 
  • #7
LeonhardEuler
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The terms aren't even approaching zero.
 
  • #8
789
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Right, but there is a cancellation.
 
  • #9
LeonhardEuler
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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
789
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:
  • #11
LeonhardEuler
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Yes, I think thats what it is. He said he needed an approximation, so do you think he should just integrate it?
 
  • #12
mathwonk
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just adding up a few terms makes it look really easy. but maybe i am too optimistic with small evidence.
 
  • #13
Curious3141
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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
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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....
 

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