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

  • Context: High School 
  • Thread starter Thread starter stupidkid
  • Start date Start date
  • Tags Tags
    Value
Click For Summary

Discussion Overview

The discussion revolves around finding the approximate value of the series 1/1*2 + 1/2*3 + 1/3*4. Participants explore methods for evaluating the series, including sigma notation and telescoping series, while also addressing convergence issues.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions if there is a simple way to find the answer to the series.
  • Another participant suggests using sigma notation and simplification techniques.
  • Some participants express confusion over the interpretation of the series, with differing views on whether it converges.
  • One participant proposes that the series can be expressed as a telescoping series, indicating potential for cancellation of terms.
  • There is a suggestion to approximate the series by integrating it, although this is met with skepticism regarding the need for approximation.
  • Another participant emphasizes that the series can be evaluated exactly using partial fraction decomposition.
  • Concerns are raised about the convergence of the series, with some arguing that the terms do not approach zero, while others mention cancellation effects.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the convergence of the series or the necessity of approximation. Multiple competing views remain regarding the interpretation and evaluation methods for the series.

Contextual Notes

There are unresolved assumptions about the participants' familiarity with mathematical concepts such as sigma notation and telescoping series, which may affect the discussion's direction.

stupidkid
Messages
18
Reaction score
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?
 
Mathematics news on Phys.org
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
 
Oh man, I thought it was 1/(1*2)+1/(2*3) ... and that was really hard. That's so much easier though.
 
Hmmm... I guess it could be like that too. The OP will probably post if my interpretation was incorrect.
 
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.
 
Doesn't converge? Would you explain why not? I see a convergence.
 
The terms aren't even approaching zero.
 
Right, but there is a cancellation.
 
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
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
Yes, I think that's what it is. He said he needed an approximation, so do you think he should just integrate it?
 
  • #12
just adding up a few terms makes it look really easy. but maybe i am too optimistic with small evidence.
 
  • #13
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
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...
 

Similar threads

High School Arithmetic and our place value system
  • · Replies 6 ·
Replies
6
Views
2K
High School Given an equation, find the value of this expression
  • · Replies 17 ·
Replies
17
Views
3K
Approximate value of ##E=1/2! +1/3!+1/4!+... ##
  • · Replies 12 ·
Replies
12
Views
3K
Undergrad 10^(1/10) is VERY close to 2^(1/3)
  • · Replies 3 ·
Replies
3
Views
1K
MHB For j >1 the first 2^j digits are correct
  • · Replies 4 ·
Replies
4
Views
1K
Undergrad Why fixed point iteration of ##x^3 = 1-x^2## doesn't converge when ##x_0= 0##
  • · Replies 16 ·
Replies
16
Views
3K
MHB [ASK] Addition of Fractions: 1/3+1/6+1/10+1/15+1/21+....+1/231
  • · Replies 3 ·
Replies
3
Views
3K
MHB Find (x₁²+1)(x₂²+1)(x₃²+1)(x₄²+1)
  • · Replies 2 ·
Replies
2
Views
2K
MHB Question: What is the value of ⌊ 2020/(1+2+3+...+2019)⌋?
  • · Replies 2 ·
Replies
2
Views
2K
MHB Divisibility of (1!+2!+3!+...+100!)^2 by 5
  • · Replies 1 ·
Replies
1
Views
1K