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I Error in Series Approximations

  1. Aug 11, 2017 #1
    It has been defined that for an alternating series, the difference between the total sum of the series and the partial sum of the series through nth term is always less than or equal to the (n+1)th term. Can anyone explain the intuitive reason behind this?
  2. jcsd
  3. Aug 11, 2017 #2
    It's an interesting "definition" - but there could be exceptions.

    For an alternating series of real values that converges, what is most commonly seen is that ending on a "plus" term puts you over the correct value and ending on a "minus" term puts you under the correct value. So you steadily close in on the correct value, but you keep crossing from one side to the other as you add each consecutive term.

    Given that scenario, you would know that the next term (N+1) is going to overshoot the correct value - and therefor your current error is less than the absolute value of that term.

    For example, consider the series ## \sum_{n=0}^{\infty}V^{n} = 1/{V+1} = V + V^2 + V^3 + ...## where ##0>V>-1##.
    For V=-0.5, the full sum ("correct value") is 2/3. The partial sums go: 1, 1/2, 3/4, 5/8, 11/16, 23/32, ... . In each case the next partial sum ends up beyond the correct value.

    Ahhh. See mfb's post below for the rest of the story.
    Last edited: Aug 11, 2017
  4. Aug 11, 2017 #3


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    Staff: Mentor

    You need some additional condition for that. ##|a_{n+1}|<|a_n|## is not necessary but sufficient to make that statement true.

    Let's assume we have this additional condition, and let's say ##a_n>0## (the proof works exactly the same with flipped signs). Then ##a_{n+1} + a_{n+2}<0## and ##a_{n+3} + a_{n+4}<0## and so on. The sum of all these pairs is negative as well, so the sum of the first n terms overestimates the limit.
    At the same time, we know that ##a_{n+2} + a_{n+3}>0## and ##a_{n+4} + a_{n+5}>0## and so on. The sum of the first n+1 terms underestimates the limit. It is ##|a_{n+1}|## smaller than the sum of the first n terms, therefore the error after the nth term is at most the size of the next term.
  5. Aug 11, 2017 #4
    Thank you for the excellent intuitive answers.
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