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I feel like something is wrong here in my logic but why do these two equations not equal each other?

Equation 1: $\sum_{i=0}^{\infty}(0)$

Steps to equation 2:

$0 = x - x$ where $x \in \mathbb{R}$

therefore,

$\sum_{i=0}^{\infty}(x - x)$

which could be represented as:

$x - x + x - x + x - x ...$

Moving the first $x$ into it's own component would yield into:

$x + \sum_{i=1}^{\infty} (-x + x)$

As $-x + x = 0$, the second term would drop which would imply $0=x$ which I'm pretty sure is wrong but not sure why. I think something is wrong in the way I'm converting $x - x + x - x + x - x ...$ into $x + \sum_{i=1}^{\infty} (-x + x)$.

EDIT:

I also realized that using the same logic as above, the resultant equation could also be expressed as:

$x - x + \sum_{i}^{\infty} (x - x + x)$

or

$x + \sum_{i}^{\infty} (- x + x - x)$

which would mean the expression I have here would oscillate between various results. I think there's definitely something wrong then in the translation of $x - x + x - x + x - x ...$ to the other equations but not sure why?

RohanP
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3 Answers3

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You've rediscovered the Eilenberg-Mazur swindle. This is a famous example showing that you can't arbitrarily manipulate the terms of a sum.

Think about it like this: in the sequence $\displaystyle\sum_{i=0}^\infty (-1)^ix$, at step $0$ you add $x$. Then at step $1$, you subtract $x$. This gives you a sequence of partial sums that goes $(x,0,x,0,...)$ which clearly does not converge, since it oscillates back and forth between two values forever. Whereas in your initial sum, you are "doing the addition and subtraction of $x$ in the same step." This results in a sequence of partial sums of the form $(0,0,0,...)$ which indeed converges to $0$.

Infinite sums are confusing because they're secretly just sequences of the partial sums. And by doing the manipulation you did, you did something illegal because you manipulated the sequence in a way that is not allowed.

Charlie
  • 665
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It has to do with the definition of an infinite sum. When we write $\sum_{n=1}^{\infty}a_n$ we mean that we are constructing a new sequence $(b_n)$ where $b_1 =a_1$, $b_2=a_1+a_2$, $b_3 = a_1+a_2+a_3$ and so on and you examine the limiting behavior of $(b_n)$.

Thus it depends on what you mean by $x - x + x - x + x - x \ldots$ If you mean $\underset{a_1}{\underbrace{x-x}} + \underset{a_2}{\underbrace{x-x}} + \underset{a_3}{\underbrace{x-x}} \ldots $, then the infinite sum is equal to $0$

Bu if you mean $\underset{a_1}{\underbrace{x}} + \underset{a_2}{\underbrace{(-x+x)}}+ \underset{a_3}{\underbrace{(-x+x)}} \ldots $ then the infinite sum is equal to $x$.

If you mean that $\underset{a_1}{\underbrace{x}} + \underset{a_2}{\underbrace{(-x)}}+ \underset{a_3}{\underbrace{x}} \ldots $ then it diverges.

All that has to do with the definition, and with different $(a_n)$ you consruct different $(b_n)$. So even though it seems you write down the same thing, it really isn't.

alphaomega
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Note that $$x - x + x - x + \dots = \sum_{i=0}^\infty (-1)^i x = x \sum_{i=0}^\infty (-1)^i $$ which is a divergent series. It is almost a problem that most people debate on without taking into note the divergence of the actual series, namely: $$1 - 1 + 1 - 1 + 1 - \dots = ?$$ While many people believe it is $0$, others "prove" it to be $1/2$ as follows

Let $S$ denote the value of the sum (which is illegal since it does not converge to a constant value). Then, we have $$\begin{align}&S = 1 - (1 - 1 + 1 - 1 + \dots) = 1 - S \\&\implies S = 1 - S \\&\implies S = \frac12\end{align}$$

VIVID
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