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Error in Series Approximation Proof

  1. Jun 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove: Capture.png
    Hint: Group the terms in the error as ##(a_{n+1}+a_{n+2})+(a_{n+3}+a_{n+4})+\cdots## to show that the error has the same sign as ##a_{n+1}##. Then group them as ##a_{n+1}+(a_{n+2}+a_{n+3})+(a_{n+4}+a_{n+5})+\cdots## to show that the error has magnitude less than ##\left| a_{n+1} \right |##.

    2. Relevant equations
    3. The attempt at a solution

    From the first part of the hint I can see that the grouped terms ##(a_{n+1}+a_{n+2})+\cdots## will take the sign of the first term because ##\lim_{n\to\infty}a_{n}=0## therefore, ##\left | \lim_{n\to\infty}a_{n+1} \right | > \left | \lim_{n\to\infty}a_{n+2} \right | ## At this point I am not really sure what to do with the second part of the hint although I'm assuming it has something to do with the less than or equal to symbol.
     
    Last edited: Jun 9, 2015
  2. jcsd
  3. Jun 9, 2015 #2

    Mark44

    Staff: Mentor

    No, that's not it. You are not using the fact that this is an alternating series, where |an + 1| < |an|.
     
  4. Jun 9, 2015 #3
    Edit: Confused myself
    Edit #2: I don't understand I thought that's what I used to show that if ##a_{n+1}>0## then ##a_{n+1}+a_{n+2}>0##
     
    Last edited: Jun 9, 2015
  5. Jun 9, 2015 #4
    Meant to edit my last post...
     
  6. Jun 9, 2015 #5

    Mark44

    Staff: Mentor

    I don't see how this helps your cause.

    If ##a_{n+1}>0##, how do ##a_{n+1}## and ##a_{n+1} + a_{n + 2}## compare? By that, I mean, can you determine which expression is larger?

    It might help your intuition to think about a specific alternating series, such as 1 - 1/2 + 1/3 - 1/4 + 1/5 -+ ..., and grouping terms both ways as suggested in the hint.
     
  7. Jun 10, 2015 #6
    Well I thought I had shown that the magnitude of ##a_{n}## was larger than ##a_{n+1}## by assuming than ##a_{n}## is positive and then stating that ##a_{n}+a_{n+1}>0##.

    I have that if the last term I include in my calculation is ##a_{n}## and ##a_{n}## is negative, and also using the fact that ##a_{1}>a_{2}>\cdots>a_{n}>a_{n+1}>a_{n+2}>\cdots>a_{k}##
    Then for the remainder, i.e. the terms omitted from the calculation, ##R_{n}=a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+\cdots+a_{k}##

    Grouping the terms like ##R_{n}=(a_{n+1}+a_{n+2})+(a_{n+3}+a_{n+4})+\cdots+(a_{k-1}+a_{k})## it can be seen that the remainder has to be a positive number since the first term in the parenthesis will always be positive and larger than the 2nd term in the parenthesis which is negative.

    Next by regrouping the terms like ##R_{n}=a_{n+1}+(a_{n+2}+a_{n+3})+(a_{n+4}+a_{n+5})+\cdots## it can be seen that after the first term ##a_{n+1}## all of the parenthesis will be negative therefore the remainder will be less than ##a_{n+1}##. So I have that ##R_{n}## is greater than 0 and less than ##a_{n+1}## if the last term included in my calculation is negative.
    ##S=S_{n}+R_{n} \Longrightarrow S-S_{n}=R_{n}## now since ##0<R_{n}<a_{n+1}## I can replace ##R_{n}## with ##a_{n+1}## and change the sign to an inequality which gives $$S-S_{n}<a_{n+1}$$ At this point I was wondering if I could just take the absolute value of both sides to finish the proof or do I have to go over the other case where the last term I calculate is positive.
     
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