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Infinity puzzle

  1. Mar 31, 2008 #1
    Tell me what is wrong with this :)

    ln (2) = ln( 1 +1 ) and the power series expansion of ln(1+x) for x=1 gives

    ln(2) = 1 - 1/2 + 1/3 -1/4 + 1/5 + ......
    = (1 + 1/2 + 1/3 + 1/4 + 1/5 + ......... )
    - 2 . ( 1/2 + 1/4 + 1/6 + ..............)

    = (1 + 1/2 + 1/3 + 1/4 + 1/5 + ......... )
    - (1 + 1/2 + 1/3 + ................)
    = 0 = ln (1)

    => 2 = 1 ! huh !
     
  2. jcsd
  3. Mar 31, 2008 #2

    HallsofIvy

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    The interval of convergence for ln(1+ x) at x= 1 is (0, 1). The series you get at x= 1, 1+ 1/2+ 1/4+ ... does not converge so your calculation is invalid.
     
  4. Mar 31, 2008 #3
    you are partially right. you are right n the sense that (1 + 1/2 + 1/3 + 1/4 ...) is divergent. but you are worng in the sense that you only considered that portion of the sum, the whole thing together is not divergent. (x + x) - 2x : is convergent even if (x+x) is divergent :)
     
  5. Mar 31, 2008 #4
    The definition of a Taylor expansion relies on the ordering of the terms. When manipulating infinite sums, it is no longer true that addition commutes. This is a well-known issue, and many brilliant minds have worked to understand it. A quick Google for "divergent sums" will get you a long way towards what people have already thought of.
     
  6. Mar 31, 2008 #5
    agreed. a more elegant answer is here :)

    ln(2) = 1 - 1/2 + 1/3 -1/4 + 1/5 - 1/6 ...... +- 1/2n where n -> inf
    = (1 + 1/2 + 1/3 + 1/4 + 1/5 + + .... 1/2n where n -> inf
    - 2 . ( 1/2 + 1/4 + 1/6 + ... this has only n terms )

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

    = 1/n+1 + 1/n+2 + 1/n+3 .... 1/2n

    = intergral of (1/1+x) from 0 to 1 , from newton's summation formula of definite integral

    = ln (1+x) from 0 to 1

    = ln 2

    lol !
     
  7. Apr 1, 2008 #6

    ssd

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    The series is not absolutely convergent. Therefore rearrangement if its terms is not permissible.
     
  8. Apr 1, 2008 #7

    Gib Z

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    The interval of convergence for ln (1+x)'s series is (0,1]. When x=1, the series is famously the "alternative harmonic series" and equal to ln 2.
     
  9. Apr 2, 2008 #8
    >> The interval of convergence for ln (1+x)'s series is (0,1].

    I thought the interval is (-1,1], open on -1, closed on +1
     
  10. Apr 2, 2008 #9

    Gib Z

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    Yup you are correct. My main point still holds though =]
     
  11. Apr 2, 2008 #10
    yeah be careful with the infinite series.
    0=+1 + (-1) right?
    so
    0= 1-1+1-1+1-1+1-1...
    = 1+(-1+1)+(-1+1)+(-1+1)...
    = 1 +0+0+0+0...
    =1

    which is a load of crap
     
  12. Apr 2, 2008 #11
    really now, everyone knows that 1 - 1 + 1 - 1 ... = 1/2

    Of course, when I write it like that, I mean ((((1 - 1) + 1) - 1) + 1) and so on

    of course (1-1) + (1-1) + (1-1) ... = 0, and nothing else. What else could it consistently be?

    Anyway, whether or not 1 - 1/2 + 1/3 - 1/4 ... converges doesn't actually matter here. The fact is that 1 + 1/2 + 1/3 + 1/4 + ... does not converge, so the first series I wrote down can't be rearranged or you can do so to get any number that you want at all.
     
  13. Apr 7, 2008 #12
    The series you construct from the first series is only half as long. Therefore you can't do it. Half the subtraction sum can't be finished.
     
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