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Proof of the Alternating Series Approximation Theorem

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Problem taken from Boas Mathematical Methods book, Section 14 page 35.

    Prove that if ##S=\sum_{n=1}^{\infty} a_n## is an alternating series with ##|a_{n+1}|<|a_n|##, and ##\lim_{n \to \infty} a_n=0##, then ##|S-(a_1+a_2+...+a_n)|\leq|a_{n+1}|##.

    3. The attempt at a solution
    Hint: Group the terms in the error as ##(a_{n+1}+a_{n+2})+(a_{n+3}+a_{n+4})+...## to show that the error has the same sign as ##a_{n+1}##

    Because alternating series, and ##|a_{n+1}|<|a_n|##
    If ##a_{n+1}>0## and ##a_{n+2}<0##, then ##(a_{n+1}+a_{n+2})>0##. Then the same thing will happen with the subsequent group for all ##n##, and the whole error will have the same sign as ##a_{n+1}##. The same can be shown if ##a_{n+1}<0##.

    Hint: Group them (the error) as ##a_{n+1}+(a_{n+2}+a_{n+3})+(a_{n+4}+a_{n+5})+...##
    To show that the error has magnitude less than ##|a_{n+1}|##.

    It is true that ##(a_{n+2}+a_{n+3})=(S_{n+3}-S_{n+1})##. It can be shown using induction that the group of the terms can be represented as the partial sum of the whole series as shown above for all##n##.

    Hence it is simple to see that because the series is convergent the partial sum will tend to ##S## and the whole group of terms beside ##a_{n+1}## will go to 0. This implies the equation in the theorem.

    Alternatively, it can be shown that because of alternating series ##(a_{n+2}+a_{n+3})+(a_{n+4}+a_{n+5})+(...)<0## because of the grouping of the term. The positive odd terms will be smaller than the negative even terms after ##a_{n+1}## and from this we can directly also get the inequality. ##|S-S_n|\leq|a_{n+1}|##

    Can you guys help me to verify my flow of thoughts and proof?

    Thank You
     
    Last edited: Oct 17, 2013
  2. jcsd
  3. Oct 17, 2013 #2
    You are correct in your statements about the 1st hint, although you don't make it as clear as you could. If ##a_{n+1} > 0## then ##a_{n+1}- a_{n+2}## > 0 and true for all subsequent pairs. So the error term will have a + sign. Same argument if ##a_{n+1}## < 0.

    I got a little lost in your presentation after that. The second hint is a good one, and I think you should show that, but I don't think you do; or else I didn't understand your presentation. I'd start something like this: If ##a_{n+1}## >0, what is the sign of ##a_{n+2} +a_{n+3}##?.

    Once you've proved that step you can say ## S - S_{n} = \sum_{k=1}^\infty a_{n+k} ## = ##a_{n+1}## + ## \sum_{k=2}^\infty a_{n+k}##. Now what do you know about that last sum?

    I don't see any need for induction in this case.

    I would like to point out that we do not know that the series is convergent, unless you refer us to a theorem that says it must be. I would say we are on the verge of proving the series is convergent, although what is here is not written quite carefully enough.
     
  4. Oct 17, 2013 #3
    Ok no problem with the first hint.

    Ok I made a mistake about expressing the term as a sum of the series because you can't just take a limit of it and prove the inequality.

    I just realized that the alternative method makes more sense because if ##a_{n+1}>0##, certainly the rest of the sum would be negative## and with it I can get the inequality in the theorem. I think the hint is very useful if you think it in terms of the sign.

    I think the book also assumes that the series meets the alternating series for convergence, and hence it's convergence.

    Thanks
     
  5. Oct 17, 2013 #4
    Yes, it is true that alternating series which has terms going to zero will converge; and this problem nearly proves it. The only problem with this as a proof is that you can't assume an S if there may not be one. But that can be worked around.
     
  6. Oct 18, 2013 #5
    Can we just assume there's always S? How do you suggest we go around it?
     
  7. Oct 18, 2013 #6
    If it doesn't converge, there wouldn't be a finite S. Better practice is to write it as
    ##\sum_{k=1}^{\infty} - \sum_{k=1}^n \le |a_{n+1}|##,
    or if you want to save writing begin by defining S = ##\sum_{k=1}^{\infty}##.

    In this case the distinction is trivial because you have that strong inequality, which implies that there must be convergence. However, I've had a lot of problems in the past with writing things sloppily and then having trouble, so I tend to be careful.
     
  8. Oct 18, 2013 #7
    Ok I understand that it really needs to be very clear.

    You're summing ##a_n## right?
     
  9. Oct 18, 2013 #8
    Yes I am summing ##a_n##, but my Latex fingers were tired. So now I am the one not being completely clear.

    Best wishes.
     
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