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Suppose we split a line which length is $1$ in half. Then we get the $2$ lines which length is $1/2$ . Divide these two lines equally in half. Suppose we repeat this process infinitely. The length of lines can be calculated $\lim_{N \to \infty}$$\frac{1}{2^N}$ .But the sum of lines doesn't get $1$.

Let $\ \lim_{N \to \infty}$$\frac{1}{2^N}=\varepsilon$, where $\varepsilon \ge 0$

if $\varepsilon > 0$ then, $\sum_{n=1} ^{\infty} \varepsilon =\infty$

and if $\varepsilon = 0$ then, $\sum_{n=1} ^{\infty} \varepsilon =0$

Why does this contradiction occur?

최민욱
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    $1 = 2\cdot \frac{1}{2}=4\cdot\frac{1}{4}=\dots = 2^n\cdot\frac{1}{2^n}$. Indeed, $\lim\limits_{n\to\infty}\left[2^n\cdot\frac{1}{2^n}\right]$ is equal to $1$. This is not to be confused with having pushed the limit inside of the expression. This is not the same thing as $\left(\lim\limits_{n\to\infty} 2^n\right) \cdot \left(\lim\limits_{n\to\infty}\frac{1}{2^n}\right)$ – JMoravitz May 19 '22 at 14:41
  • @Balloon I don't know why you bring up $\sum\limits_{n=1}^\infty\frac{1}{2^n}=1$ when the OP is asking about $1 = \frac{1}{2}+\frac{1}{2} = \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\underbrace{\frac{1}{8}+\frac{1}{8}+\dots+\frac{1}{8}}_{8~\text{copies}}$. Their argument is that as we look at the limit of this process we are essentially adding "infinitely many infinitesimals" which should either be infinity if the infinitesimal is positive or zero if the infinitesimal is zero. The punchline, again, is that the size of the summand is explicitly tied to the number of terms. – JMoravitz May 19 '22 at 14:47
  • ... and that the limit of the summand can not be taken separately with the limit of the number of terms in the sum. They must both vary simultaneously as a part of the same expression, not separately. – JMoravitz May 19 '22 at 14:48
  • I understand thank you. – 최민욱 May 19 '22 at 15:14

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