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Is the sum of this infinite series 0 or 1?

  1. Jan 13, 2012 #1
    1 - 1 + 1 -1 + 1 - 1 ...

    Does that equal 0 or 1?

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

    or

    1 + (-1 + 1) + (-1+1) + (-1+1) + ... = 1 + 0 + 0 + 0 = 1
     
  2. jcsd
  3. Jan 13, 2012 #2

    disregardthat

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    For a series to converge it must "tend to" or "approach" a number [itex]s[/itex]. Mathematically there must be a natural number [itex]N[/itex] for every real number [itex]\epsilon > 0[/itex] such that [itex]|s_n-s| < \epsilon[/itex] for [itex]n \geq N[/itex], where [itex]s_n[/itex] is the partial sum [itex]\sum^n_{k=1} a_k[/itex]. In this case we say that the sum converges to [itex]s[/itex].

    Note that it is important what you define as terms (the [itex]a_k[/itex]'s are the terms). 1-1+1-1+1-... is not the same as (1-1) + (1-1) + ...

    The first sum has terms 1,-1,1,-1,... and so on, but the second sum has 1-1,1-1,1-1,... that is, 0,0,0,... as terms.

    Obviously, if 1-1 = 0 are the terms, the series will converge to 0. The sum is simply 0 + 0 + 0 + ... which converges to 0 in the mathematical sense described above. Your example was 1 + (1-1) + (1-1) + ... which of course converges to 1. But this is not the same series.

    On the other hand, if the terms are 1, -1, 1, -1, ... the series does not converge. And it is still not the same series as the other one. The reason is that it doesn't "tend to" or "approach" any number. The partial sum [itex]s_n[/itex]oscillates between 1 and 0. Mathematically, it doesn't satisfy the condition for convergence to any number [itex]s[/itex] when we choose [itex]\epsilon = \frac{1}{2}[/itex]. Can you see why?

    Be careful around infinite series. Customary properties such as associativity and commutativity of terms doesn't apply in the same way.
     
  4. Jan 13, 2012 #3


    So is his method for finding the sum of that infinite series incorrect?

    Thanks for clearing that up by the way.
     
    Last edited by a moderator: Sep 25, 2014
  5. Jan 13, 2012 #4

    D H

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    This is Grandi's series. One answer: It's anything you want. It's an alternating series that is not absolutely convergent. You have found two arrangements that give two different answers. You can rearrange the terms to give you any number whatsoever. Another way to put this: It's indeterminate.

    However, if you insist on assigning a value to this series, the best such value is 1/2. Beware: The techniques used to do this will also say that 1+1+1+1+... = -1/2 and that 1+2+4+8+...=-1.
     
  6. Jan 13, 2012 #5

    disregardthat

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    Technically it is incorrect using the definition of convergence I described above. The series 1+2+4+16+... does not converge (it diverges), and summing two infinite series require convergence of both.

    However there are other kinds of summations, see

    http://en.wikipedia.org/wiki/1_+_2_+_4_+_8_+_…

    which is something we customarily don't use when summing series. But in that context it can in fact be so that 1 + 2 + 4 + 16 + ... = -1. But that's not to say that the series "approach" or "tend to" -1. The methods used in the video are strictly incorrect though.
     
    Last edited by a moderator: Sep 25, 2014
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