# How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots=0$?

Let

\begin{align*}x &= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} + \cdots\\ y &= \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots\\ z &= \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots \end{align*}

so we have

$$x = y + z.$$

However, $x = 2\cdot z$, so $y$ = $z$ or

$$\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots$$

This looks ok if I interpret it as

$$\frac{1}{1} = \left (\frac{1}{2} - \frac{1}{3} \right ) + \left (\frac{1}{4} - \frac{1}{5} \right ) + \left (\frac{1}{6} - \frac{1}{7} \right ) + \cdots + \left (\frac{1}{2n} - \frac{1}{2n+1} \right ) + \cdots$$

However, it's a bit weird if I write it as

$$\left (\frac{1}{1} - \frac{1}{2} \right ) + \left (\frac{1}{3} - \frac{1}{4} \right ) + \left (\frac{1}{5} - \frac{1}{6} \right ) + \cdots + \left (\frac{1}{2n-1} - \frac{1}{2n} \right ) + \cdots = 0.$$

How can a sum of positive numbers equal $0$?

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Maybe the series in question does not converge uniformly absolutely, so that you cannot arrange terms as you wish? In fact, I do not think it converges at all... –  awllower Jun 14 '13 at 7:23
@awllower: "uniformly" doesn't apply to this context, and "absolutely" is redundant because all terms in the orginal series are positive. –  Jonas Meyer Jun 14 '13 at 7:34
@athos: Please give your questions descriptive, informative titles (so that people can understand what the question is about from the title). –  Zev Chonoles Jun 14 '13 at 7:56
This is called Riemann's Theorem: a conditionally convergent infinite series can be rearranged so the series converges to any given value. –  Eric Jablow Jun 14 '13 at 11:03
@JonasMeyer: Thanks for your reminder. Now I wonder which term should be used... Noitce that the series has negative terms, as OP subtracts some terms. –  awllower Jun 15 '13 at 5:45

Almost everything in your proof works fine until you write "this looks ok if I interprete it as..."

Until then, you are manipulating infinite sums of series with positive terms, these are extended nonnegative real numbers (numbers $x$ such that $0\leqslant x\leqslant+\infty$, if you like) hence adding them and equating them is perfectly legal.

The trouble begins when you substract them, since there is no substraction on the set of extended nonnegative real numbers. Unsurprisingly, you soon must deal with $(+\infty)-(+\infty)$ differences, and chaos ensues.

A less sophisticated example, flawed quite similarly, is to start with the correct identity $$1+1+1+\cdots=\underline{\mathbf 1}+(\color{red}{1}+\color{blue}{1}+\color{green}{1}+\cdots),$$ and to deduce from it that $$0=(1-\color{red}{1})+(1-\color{blue}{1})+(1-\color{green}{1})+\cdots=\underline{\mathbf 1}.$$

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Not quite everything is fine before "this looks ok..." For extended nonnegative real numbers $x,y,z$, from $x=y+z$ and $x=2z$ one cannot deduce in general that $y=z$. (One can in this case, but there is a gap in the reasoning.) –  Jonas Meyer Jun 14 '13 at 7:45
@JonasMeyer Yes. –  Did Jun 14 '13 at 10:05
@Did thanks, now i got it clearly :) –  athos Jun 17 '13 at 2:43

Your proof is not right since $x=\infty$.

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so... basically when talking about infinite numbers, "=" shall not be used? –  athos Jun 14 '13 at 7:37
@athos: Actually "minus" should generally not be used (or its use must be justified). –  Najib Idrissi Jun 14 '13 at 8:02

Answering only "How can a sum of only positive numbers equal 0?" : look at $1^2 + 2^2 + 3^2 + 4^2 + 5^2+ ... = 0$ which is a meaningful result if the series is understood as zeta-series.

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You are manipulating divergent series. When you do the substraction, you can in fact obtain any limit you want by changing the order in which you are adding or subtracting things.

For example, $(1 - \frac 12 - \frac 14 - \frac 18 - \ldots) + (\frac 13 - \frac 16 - \frac 1{12} - \frac 1{24}- \ldots) + (\frac 15 - \frac 1{10} - \frac 1{20} - \frac 1{40} - \ldots) + \ldots = 0 + 0 + 0 + \ldots = 0$

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