- #1

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

Is there a simple way to find the anwer to problems like this?????

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- Thread starter stupidkid
- Start date

- #1

- 18

- 0

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

Is there a simple way to find the anwer to problems like this?????

- #2

- 644

- 1

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

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

- #4

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- #5

Zurtex

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

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

Does clearly not converge.

- #6

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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

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Right, but there is a cancellation.

- #9

LeonhardEuler

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- #10

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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?

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

Correct?

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- #11

LeonhardEuler

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- #12

mathwonk

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- #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.

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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If the sum is:mathwonk said:

[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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