Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Alternating Series Approximation - Please help

  1. Apr 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine the number of terms required to approximate the sum of the series with an error of less than .001

    Sum ((-1)^(n+1))/(n^3) from n=1 to infinity

    2. Relevant equations

    3. The attempt at a solution

    I guess this is what you do:

    1/(n+1)^3 < 1/1000

    and solving you get n+1 > 10 so the answer is 10 terms

    But that doesn't quite make sense to me, and I'm not sure why.

    Alternating series remainder theorem:

    |S-Sn| =|Rn|< or = to an+1

    Could someone please explain this to me?
  2. jcsd
  3. Apr 9, 2008 #2
    Using 1/(n+1)^3 < 1/1000 will not yield the right result. See the below logic.

    An alternating series alternates between positive and negative terms, correct? If you look at an n-term approximation of the sum of an alternating series, you expect that n-term approximation to alternate between "overshooting" and "undershooting" the total (infinite term) sum, with increasing convergence toward the actual infinite-term summation. Since adding a subsequent term to the summation always takes the sum from "undershooting" to "overshooting" or vice versa, the magnitude of the n+1th term of the sequence is always greater than the error between the n-term sum and the infinite term sum. This is what your remainder theorem says, in words.

    The nice part about the theorem is that it gives a way to determine an appropriate approximation without knowing the actual sum. You're given an Rn=0.001, now simply find the smallest a_{n+1} satisfying a_{n+1}>=1/1000. Note if you were to (for example) choose the second smallest a_{n+1}, your error term S-Sn would yield too large an error.
    Last edited: Apr 9, 2008
  4. Apr 9, 2008 #3
    Yeah, I see what you're saying and that's why I'm confused. The logic 1/(n+1)^3 < 1/1000 is what my solutions manual says. Then they go on to say that n+1 > 10 and that you need 10 terms.

    What would you say that the correct answer is?
  5. Apr 9, 2008 #4
    Solutions manuals aren't right all the time. Indeed, the difference between the solutions manual and the theorem is 1 term (9 terms is the right answer), but the requirement in the solutions manual is tighter than needed and fails to take into account the nuances of alternating series.
    Last edited: Apr 9, 2008
  6. Apr 9, 2008 #5
    Actually, I've been doing some thinking, and I think that the manual is right.

    According the manual's logic, the an+1 term is less than 1/1000, and according to the theorem, our error (the remainder) is less than or equal to the an+1 term. So if an+1 is less than 1/1000 then the error must also be less than 1/1000. So I think that if you find n so that 1/(n+1)^3 = 1/1000 (n=9) and then go up to the next integer so that you satisfy the inequality, 10, that is the correct # of terms.
  7. Apr 9, 2008 #6
    It's definitely a subtle point, but my answer is vindicated by actually computing the sequence. The sequence converges to 0.9015426... (and so on), and the 9-term summation is 0.90207, which is clearly within the error margin of 1/1000.

    Correct, but if a_{n+1} is NOT less than 1/1000, the error isn't necessarily greater than 1/1000 because Sn+a_{n+1}-S is of the opposite sign as Sn-S, so some fraction of a_{n+1} added to Sn is equal to S. Let's say a_{n+1}=x+y, where x is the aforementioned fraction, then Sn+x=S. It's evident that the chunk "y" is the only part of a_{n+1} that actually contributes to summation error since Sn+x=S, and therefore Sn+x+y=S+y. Your solutions manual assumes that all of a_{n+1} contributes to error, which is untrue as I have attempted to explain.

    If this argument fails, it's ASCII art time :P
    Last edited: Apr 9, 2008
  8. Apr 9, 2008 #7
    Wow. I think I understand your argument. This is some pretty deep thinking. I really appreciate your help. Thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook