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Convergent series

  1. Aug 7, 2013 #1
    In my book it says that the series (1+x)^n converges for x<1.

    However I put n = -1 and wolfram says that the series does not converge.

    However if I let x = 1/y where y>1

    then the expansion of (1+1/y)^-1 is equal to: (which I will define as (SERIES 1))
    1 - y + (1/y)2 - (1/y)3 + (1/y)4 - ......

    = 1 + ( (1/y)3 + (1/y)5 + ... ) - ( (1/y)2 + (1/y)4 + ...)

    The series (1/y)3 + (1/y)5 + ... (1/y)3 + 2n + .. is equal to (which I will define as (SERIES 2))
    Ʃ(1/y)3 + 2n where n→∞ and 1≤n<∞.
    I also know that the sum of the (SERIES 2) is less than the series 1 + 1/4 + 1/8 + 1/16 + ... which is convergent therefore (SERIES 2) is convergent.

    Also from (SERIES 1) I know that the sum is positive and therefore the series
    ( (1/y)2 + (1/y)4 + ...) is less than (SERIES 2) + 1 and therefore as the number of terms approaches ∞ the series
    ( (1/y)2 + (1/y)4 + ...) which is positive is less than a [(finite number) + 1] which is a finite number and therefore is convergent.

    ∴Therefore the (SERIES 1) is convergent.

    Am I and the book wrong or is wolfram?

    Attached Files:

  2. jcsd
  3. Aug 7, 2013 #2


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    Staff: Mentor

    (1+x)^n is not a series.

    If you want to sum that over natural n, it is pointless to set n to anything special (in particular, negative values).

    Do you want to get the taylor expansion at x=0? That converges for x<1 for all n, and for all x for integer n (including 0 if we define 00=1).

    That step requires absolute convergence, so
    1 + |1/y| + |(1/y)2| + |(1/y)3| ......
    has to converge, too.

    That is true for some y only.

    I would expect that the book and WolframAlpha are right. What did you use as query for WA?
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