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Convergence confirmation

  1. Jul 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Verify that the infinite series converges


    2. Relevant equations

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

    3. The attempt at a solution
    So I did just that and I got
    1/n -1/(n+1)

    So I thought to take the limit of this as n goes to infinity I got 0 as the limit. My book took the limit as n goes to infinity of 1 - 1/(n+1) which they say is equal to 1. Which I don't disagree with I don't know how they got from what I got ( and them) 1/n -1/(n+1) to 1 - 1/(n+1)
    For a look at what they did see http://www.calcchat.com/book/Calculus-ETF-5e/

    You have to put in the section and question though. It is section 9.2 question 29

    Have a look I'm lost thanks
     
  2. jcsd
  3. Jul 2, 2013 #2

    CAF123

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    Do you have to actually find the limit or just show the convergence? Consider the comparison test if is the latter. If it is the former, they used partial sums to obtain ##1 - 1/(n+1)## from which they extracted the limit.
     
  4. Jul 2, 2013 #3
    They found the limit. When you do partial fractions you get 1/n -1/(n+1)
    Which they get too. But they say that 1/n -1/(n+1) = ##1 - 1/(n+1)## Which I don't understand.


    The step between those two.
     
  5. Jul 2, 2013 #4

    CAF123

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    No, they are summing together ##n## partial sums to obtain ##S_n = 1 - \frac{1}{n+1}##. They then take the limit of partial sums (as the number of partial of sums tends to infinity) which is equivalent to summation you are looking for.
     
  6. Jul 2, 2013 #5

    LCKurtz

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    $$S_n =\frac 1 {1\cdot 2} + \frac 1 {2\cdot 3} + \frac 1 {3\cdot 4} +\dots+ \frac 1 {n \cdot (n+1)}$$ $$
    =(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+\dots +(\frac 1 n - \frac 1 {n+1})$$What happens to that second expression when you simplify it?
     
    Last edited by a moderator: Jul 2, 2013
  7. Jul 2, 2013 #6
    The second expression is just the first expression when you simplify it.

    I'm confused because I don't understand where this came from 1- (1/(n+1))
    Why do they take the limit of that and not the limit of 1/n - 1/(n+1)
    This is my question. I wanted to take the limit of the latter. Why must you do the former and what is the reason they did it this way.
     
    Last edited by a moderator: Jul 2, 2013
  8. Jul 2, 2013 #7

    Mark44

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    That's not what LCKurtz asked. What happens to the 2nd expression when you simplify it?
     
  9. Jul 2, 2013 #8
    I just answered that. I said 1/n(n+1) = 1/n -(1/n+1) Put a common denominator and the n's in the numerator cancel.
     
  10. Jul 2, 2013 #9

    LCKurtz

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    Don't just go backwards. What happens when you remove the parentheses?
     
  11. Jul 2, 2013 #10

    Mark44

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    You're not looking at the entire expressions - you're just looking at the last term in each expression. The question is - what happens when you simplify this?
    $$ (\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+\dots +(\frac 1 n - \frac 1 {n+1})$$
     
    Last edited: Jul 2, 2013
  12. Jul 2, 2013 #11
    You get 1/2 +...1/n -1/(n+1)
     
  13. Jul 2, 2013 #12

    Mark44

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    I don't. How did you get it?
     
  14. Jul 2, 2013 #13
    $$ (\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 2)+\dots +(\frac 1 n - \frac 1 {n+1})$$
    Two 1/2 dissapear
    = $$ (\frac 1 1 - \frac 1 3)+(\frac 1 3 - \frac 1 2)+\dots +(\frac 1 n - \frac 1 {n+1})$$

    Two 1/3 dissapear

    =$$ (\frac 1 1 - \frac 1 2)+\dots +(\frac 1 n - \frac 1 {n+1})$$

    $$(\frac 2 2 - \frac 1 2 + \dots +(\frac 1 n - \frac 1 {n+1})$$

    So you get what I said right?
     
  15. Jul 2, 2013 #14

    Mark44

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    There was a mistake in what LCKurtz wrote, that I didn't notice when I copied and pasted it. It should have been as below. The corrected term is 1/3 - 1/4
     
  16. Jul 2, 2013 #15
    3/4 + 1/n - (1/n+1)

    right? How does this help?
     
  17. Jul 2, 2013 #16

    Mark44

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    No.
    $$ \frac 1 1 - \frac 1 2+\frac 1 2 - \frac 1 3+\frac 1 3 - \frac 1 4+\dots +\frac 1 n - \frac 1 {n+1}$$

    I have removed all of the parentheses. When you simplify the above, what are you left with?
     
  18. Jul 2, 2013 #17
    The1/2 goes with the -1/2 and the 1/3 goes with the -1/3 leaving the 1/1 and the -1/4 combining those is 3/4 and you have 1/n -1/n+1 left like I said before. How is that wrong?
     
  19. Jul 2, 2013 #18

    Mark44

    Staff: Mentor

    You aren't getting the pattern. Here it is with some more terms.
    $$(\frac 1 1 - \frac 1 2)+(\frac 1 2 - \frac 1 3)+(\frac 1 3 - \frac 1 4)+(\frac 1 4 - \frac 1 5) + (\frac 15 - \frac 1 6) + \dots +(\frac 1 n - \frac 1 {n+1})$$

    This really isn't complicated. Possibly you are missing the significance of "..." which means "continues in the same fashion."
     
  20. Jul 2, 2013 #19
    Yeah I feel like this is getting away from my original question. I get the pattern it just keeps going that is that 1/n - 1/n+1 is telling me in general it continues this way. I just want to know why they take the limit of 1 - 1/(n+1)

    and not the limit of 1/n - 1/n+1
     
  21. Jul 2, 2013 #20

    Mark44

    Staff: Mentor

    I don't think you get the pattern. What's the term that comes before (1/n - 1/(n + 1))?
     
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