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What is wrong with the following calculation using infinite series?

  1. Jul 15, 2011 #1
    1. The problem statement, all variables and given/known data

    What is wrong with the following calculation using infinite series?

    0 = 0 + 0 + 0 + ...
    0 = (1 - 1) + (1 - 1) + (1 - 1) + ...
    0 = 1 - 1 + 1 - 1 + 1 - 1 + ...
    0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ...
    0 = 1 + 0 + 0 + 0 + ...
    0 = 1

    2. Relevant equations

    None.

    3. The attempt at a solution

    Beginning on line 4, or

    0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ...

    an error is made. Specifically, the repeating number 0 was not completely repeated. More specifically, the -1 number of the 1 - 1 pair to make zero was not inserted, thereby making this statement invalid.

    Do I have the correct idea and am I missing anything? Also, any terminology I could have used is more than welcomed!

    Much appreciation!
     
  2. jcsd
  3. Jul 15, 2011 #2

    jambaugh

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    When you write a valid equation, both sides need to be well defined. When is the value of an infinite series well defined?
     
  4. Jul 17, 2011 #3
    I think that the value of an infinite series is well defined when the partial sum Sn is close as it can to the limit L be adding infinitely many terms in the series (the sum).

    So, if I took the limit of both sides of the equation (zero and zero), both sides will be equal to zero (the limit of a constant is the constant).

    However, I do not believe I am providing the necessary statements.
     
  5. Jul 17, 2011 #4
    The reason this doesn't work is that (1 - 1) + (1 - 1) + (1 - 1) + ..., which is a type of series called an Alternating Series, does not converge (meaning it doesn't add up to a finite number) and diverging series do not conform to the normal rules of arithmetic. An alternating series is a series in the form
    [tex]\sum_{n=0}^{\infty} (-1)^n a_n[/tex]
    The conditions for this series converging are:
    [tex]\lim_{n\to \infty} a_n = 0[/tex]
    and
    [tex]a_n < a_{n-1} \;\; \forall n[/tex]

    Since [itex]a_n = 1[/itex], this series would be
    [tex]\sum_{n=0}^{\infty} (-1)^n = 1-1+1-1+\cdots[/tex]

    This series fails the first test.
    [tex]\lim_{n\to \infty} 1 = 1 \neq 0[/tex]

    So it diverges.

    This is actually a well known divergent series, known as http://en.wikipedia.org/wiki/Grandi%27s_series" [Broken].
     
    Last edited by a moderator: May 5, 2017
  6. Jul 18, 2011 #5

    jambaugh

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    Exactly, so now apply the same definition to the sum of 1's of alternating signs and as mentioned it does not converge so it is neither equal to zero nor equal to the series of zeros.
     
  7. Jul 18, 2011 #6

    Hurkyl

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    A bit of a nitpick, but this series is perfectly fine -- it is an infinite sum whose terms are all zero.

    You're thinking of the next line:
    1 - 1 + 1 - 1 + 1 - 1 + ...​
     
  8. Jul 18, 2011 #7

    Ray Vickson

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    We clearly don't need [itex]a_n < a_{n-1} \;\; \forall n[/itex], because what happens in the first N terms (for fixed, finite N) will not affect convergence.

    RGV
     
    Last edited by a moderator: May 5, 2017
  9. Jul 18, 2011 #8
    Okay, I think I have a good idea of the problem and solution, however some of the replies were a bit over my current (basic) understanding of infinite series.

    Thank you for everyone that replied!
     
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