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Math Help - Sum of 2^n can = -1?

  1. #1
    Senior Member
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    Sum of 2^n can = -1?

    My friend brought up an interesting argument that

    \sum_{n=0}^\infty 2^n=-1

    \sum^\infty_{n=0} 2^n=1+2+4+8+16+32+...

    2 \cdot \sum^\infty_{n=0} 2^n=2+4+8+16+32+64+...

    2 \cdot \sum^\infty_{n=0} 2^n - \sum^\infty_{n=0}2^n=-1

    This is obviously wrong, but what is wrong with it?
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  2. #2
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    Hello, chengbin!

    This is a classic fallacy . . .


    \begin{array}{cccccc}\text{We have the sum:} & S & =&  1 + 2 + 4 + 8 + 16 + \hdots & [1] \\<br /> <br />
\text{Multiply by 2:} & 2S &=& \quad\;\;\;2 + 4 + 8 + 16 + \hdots & [2]\\ \\[-3mm]<br /> <br />
\text{Subtract [2] - [1]:} & S &=& -1\qquad\qquad\qquad\qquad\qquad \end{array}


    The fallacy is that S = \infty and 2S = \infty.

    . . And we want: . 2S - S \:=\: \infty - \infty


    But \infty - \infty is an indeterminate form

    . . which can equal any value!


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Here's another silly example: . 1 + 2 + 3 + 4 + \hdots \;=\;-\infty


    \begin{array}{ccccccc}<br />
\text{The sum is:} & S & = & 1 + 2 + 3 + 4 + 5 + 6 + \hdots & [1] \\<br />
\text{Multiply by 2:} & 2S & = & \qquad 2 \;\;\; +\;\;\; 4 \;\;\;+\;\;\;  6 \;+\;\hdots & [2] \\ \\[-3mm]<br />
\text{Subtract [2] - [1]:} & S &=& -1 \;\;\;-\;\;\; 3 \;\;\;-\;\;\; 5 \;\;-\; \hdots \quad\;\; \end{array}


    . . Therefore: . S \;=\;-\infty



    [The sum of all the natural numbers is a negative number?]
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  3. #3
    MHF Contributor

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    The fundamental difficulty, of course, is that all that manipulation assume the thing being manipulated exists! And \sum_{n= 0}^\infty 2^n does NOT exist- that sum is not convergent.
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