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I have constructed a proof of $0 = \infty$ that I know is incorrect, although I'm not quite sure why. it goes like this: $$0 = 0 = (1-1) + (1-1) + (1-1) + ...$$

but it is also true that $0. \overline{9} = 1$ because if you let: $$n = 0. \overline{9} \implies 10n = 9. \overline{9} \implies 10n - n = 9n = 9 $$ $$\therefore n = 0. \overline{9} = 1$$

so:$$0 = (1-0. \overline{9}) + (1- 0. \overline{9}) + ...$$

and if $$0. \overline{9} = \lim_{n\to \infty} \sum_{i = 1}^n 9\left( \frac{1}{10} \right)^i$$

$$1 - 0. \overline{9} = 1- \lim_{n\to \infty} \sum_{i = 1}^n 9\left( \frac{1}{10} \right)^i = \lim_{n\to \infty} \left( \frac{1}{10} \right)^n$$

therefore it is true that:

$$0 = \lim_{n\to \infty} \left( \frac{1}{10} \right)^n + \lim_{n\to \infty} \left( \frac{1}{10} \right)^n + ...$$

$$0 = \lim_{k\to \infty} \sum_{l = 0}^k \left( \lim_{n\to \infty} \left( \frac{1}{10} \right)^n \right)$$

Call $ \left( \frac{1}{10} \right)^n = \Delta x$

$$\lim_{n\to \infty} \left( \frac{1}{10} \right)^n = \lim_{n\to \infty} \Delta x \to dx \to 0$$

$$\lim_{k\to \infty} \sum_{l = 0}^k \left( \lim_{n\to \infty} \left( \frac{1}{10} \right)^n \right) = \lim_{k\to \infty} \int_0^k dx = \lim_{k\to \infty} k$$

Therefore:

$$0 = \lim_{k\to \infty} k \to \infty$$

Q.E.Done.

So my question is this; at what step did I break math?

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5  
$$\lim_{k\to \infty} \sum_{l = 0}^k \left( \lim_{n\to \infty} \left( \frac{1}{10} \right)^n \right) = \lim_{k\to \infty} \int_0^k dx$$ This is the place where you math went wrong. –  user17762 Jan 19 '13 at 3:05
1  
$$\lim_{n \to \infty} \dfrac1{10^n} = 0$$ –  user17762 Jan 19 '13 at 3:13
1  
Nope. $$\dfrac1{10^n} \neq 0$$ but $$\lim_{n \to \infty} \dfrac1{10^n} = 0$$ –  user17762 Jan 19 '13 at 3:14
12  
If you wanted to say that you want to swap limits, which is what I assume you want to do, you cannot swap limit always.$$0 = \color{red}{\underbrace{\overbrace{\lim_{k \to \infty} \sum_{l=0}^k \left( \lim_{n \to \infty} \dfrac1{10^n}\right) = \lim_{n \to \infty} \left(\lim_{k \to \infty} \sum_{l=0}^k \left( \dfrac1{10^n}\right) \right)}^{\text{You cannot interchange limits arbitrarily}}}_{\text{This is incorrect}}} = \lim_{n \to \infty} \lim_{k \to \infty} \left( \dfrac{k}{10^n}\right) = \infty$$ –  user17762 Jan 19 '13 at 3:21
3  
I'd stop you right at $0=(1-1)+(1-1)+\cdots$ –  Pedro Tamaroff Feb 22 '13 at 19:23
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1 Answer 1

$$\lim_{k \to \infty} \sum_{l=0}^k \left(\lim_{n \to \infty} \left(\dfrac1{10} \right)^n\right) \color{red}{\neq} \lim_{k \to \infty} \int_0^{k} dx$$ This is the place where you made the mistake since $$\lim_{n \to \infty} \dfrac1{10^n} = 0$$ Note that $\dfrac1{10^n} \neq 0$ but $$\lim_{n \to \infty} \dfrac1{10^n} = 0$$ Also, the other problem is, in general, you cannot swap limits arbitrarily. $$0 = \lim_{k \to \infty} \sum_{l=0}^k \,0 = \color{red}{\overbrace{\underbrace{\lim_{k \to \infty} \sum_{l=0}^k \left(\lim_{n \to \infty} \dfrac1{10^n}\right) = \lim_{n \to \infty} \left(\lim_{k \to \infty} \sum_{l=0}^k \dfrac1{10^n}\right)}_{\text{This is incorrect}}}^{\text{You cannot interchange limits arbitrarily}}} = \lim_{n \to \infty} \lim_{k \to \infty} \dfrac{k}{10^n} = \lim_{n \to \infty} \infty = \infty$$

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