$\frac{1}{2} = \frac{1+1+1+\ldots}{2+2+2+\ldots} = \frac{(1+1)+(1+1)+\ldots}{2+2+2+\ldots} = 1$ says user Karolis Juodelė. Why does it get 7 votes up for this comment? I guess that 1/2 is not 1. See: Whats infinity divided by infinity?

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    $\begingroup$ This involves some playing fast and loose with limit processes that is not in-bounds. $\endgroup$ – ncmathsadist May 20 '13 at 0:11
  • $\begingroup$ And you don't need to mention the limit explicitly anywhere? $\endgroup$ – user8005 May 20 '13 at 0:12
  • $\begingroup$ An infinite sum is automatically a limit process. $\endgroup$ – ncmathsadist May 20 '13 at 0:13
  • $\begingroup$ That user was doing a proof by contradiction. She is saying, assume that this division is valid. Then it comes out to both 1/2 and 1. That is a contradiction and therefore the division is invalid. $\endgroup$ – Mark May 20 '13 at 0:16
  • $\begingroup$ It wasn't an answer. it was a comment. Upvoting comments isn't the same as upvoting an answer. $\endgroup$ – Thomas Andrews May 20 '13 at 0:24

I upvoted that comment because it appeared to me that it was a very clear and succinct demonstration of exactly why we can't handle expressions like $$\frac{1+1+1+\cdots}{2+2+2+\cdots}$$ the same way we can handle finite expressions. The original question asked why $\infty\over\infty$ is not simply taken to be 1; Karolis's answer shows why we do not.

  • $\begingroup$ I guess this can't be correct if we have an equal amount of 1's and 2's?: $\frac{1+1+1+\ldots}{2+2+2+\ldots} = \frac{(1+1)+(1+1)+\ldots}{2+2+2+\ldots}$ $\endgroup$ – user8005 May 20 '13 at 0:19
  • $\begingroup$ It isn't correct regardless. You can't handle expressions like that the same way you handle finite expressions, and you can't understand them in the same way either. $\endgroup$ – MJD May 20 '13 at 0:21
  • $\begingroup$ If the answer isn't 1/2, Then, why should this answer by Chindea be correct? $$ \frac{1 + 1 + \cdots}{2 + 2 + \cdots} = \lim_{n \to \infty} \frac{n}{2n} = \frac{1}{2} $$ $\endgroup$ – user8005 May 20 '13 at 0:24
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    $\begingroup$ Filip did not say that it was equal to $\frac12$. He said that even if one were to adopt a certain obvious-seeming rule for assigning values infinite quotients, one would still get inconsistent results. $\endgroup$ – MJD May 20 '13 at 0:29

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