Thread: Sum of 2^n can = -1?

1. Sum of 2^n can = -1?

My friend brought up an interesting argument that

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

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

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

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

This is obviously wrong, but what is wrong with it?

2. Hello, chengbin!

This is a classic fallacy . . .

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

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

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

But $\displaystyle \infty - \infty$ is an indeterminate form

. . which can equal any value!

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

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

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

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

[The sum of all the natural numbers is a negative number?]

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