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Problem with summing a divergent series

  1. Jun 23, 2014 #1
    Hello,

    The point of this thread is to find the mathematical error in summing divergent series. For example the series: 1+2+4+8+16+32+64+...+... (doubling the numbers, or alternatively: increasing powers of 2).

    I've seen the argument that you multiply the series by 1, then substitute (2-1) for 1. All of the terms cancel except for -1 (the series appears to add to -1).

    Now, I tried the same thing but by substituting (3-2) for 1 instead of (2-1), and the series still tends to infinity, which leads me to believe the (-1) of the (2-1) is somehow special.

    But also, I noticed that two infinite power series can be multiplied. For example, the series for e^x and the series for sin(x) can be multiplied term wise to give the series for sin(x)e^x.

    So my two questions are:

    1.) Where is the mathematical fault in the procedure first described? I originally thought you can't multiply a scalar by an infinite series, but it seems that if you can multiply two infinite series you ought to be able to multiply a scalar by an infinite series.

    2.) Also, why does substituting (3-2) instead of (2-1) change the result??? Thanks a ton in advance,

    Lee
     
  2. jcsd
  3. Jun 23, 2014 #2
    1) The rule ##\sum a_n-\sum b_n=\sum(a_n-b_n)##, which is used here to cancel terms, is only necessarily true when both ##\sum a_n## and ##\sum b_n## are convergent.

    2) Standard rules of algebra aren't necessarily true for divergent series.
     
  4. Jun 23, 2014 #3
    You definitely can multiply an infinite series by a constant, but I am guessing you aren't cancelling the terms correctly. Consider just the sum going to N, and you can take the limit N→∞ afterwards:

    [tex]
    \sum_{i=0}^{N}2^i = (2-1)\sum_{i=0}^N 2^i = \left(2 + 4 + 8 + \cdots + \cdots 2\cdot 2^N\right) -\left(1+2+4+\cdots + 2^N\right)=2\cdot 2^N -1
    [/tex]

    which you can see is still divergent.
     
  5. Jun 23, 2014 #4
    That was exactly what I was looking for. Thank you!
     
  6. Jul 2, 2014 #5
    So this method can only be used in summing a convergent, geometric series?

    [itex]S=\displaystyle\large\sum\limits_{i=0}^{\infty} 2^i = 2S+1[/itex]

    Then [itex]S=-1[/itex]
     
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