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I'm a high-school math student, and I recently stumbled upon an interesting, but counterintuitive result while solving a problem. I was trying to prove that the sum $S$ of all $x$ such that $x\in(0,1)$ was not finite. While doing so, I found this:

Using divide-and-conquer approach, $$\begin{aligned} S &= \left( \frac{1}{2} \right) + \left( \frac{1}{4} + \frac{3}{4} \right) + \left( \frac{1}{8} + \frac{3}{8} + \frac{5}{8} + \frac{7}{8} \right) + ...\infty \\ &= \sum_{i=1}^{\infty} \sum_{j=1}^{2^{i-1}} \frac{2j-1}{2^i} \\ &= \sum_{i=1}^{\infty} \frac{(2^{i-1})^2}{2^i} \\ &= \sum_{i=1}^{\infty} 2^{i-2} \\ &= \frac{1}{2} + 1 +2 + 4 + 8 +...\infty \end{aligned} $$ Is this valid or is there a flaw that I'm missing?

Upayan De
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  • I am aware of that. However, isn't this a geometric series? I understand that both the series are divergent and consequently, their summation tends to infinity. My question here is that can a summation of continuous elements be seen equivalently as the sum of distinct elements? – Upayan De May 08 '21 at 04:48
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    One thorn is how to define the sum of uncountably many addends. But your argument is fine: since the numbers are nonnegative, any reasonable definition of the full sum should be larger than the sum over a countable subset, and you've found one such subset that diverges. – angryavian May 08 '21 at 04:52
  • Well, in that all these infinite sums diverge to infinity, I guess you could say this is correct. – Thomas Andrews May 08 '21 at 05:10
  • I’m not sure I know how you are using the term “continuous element” here. – Thomas Andrews May 08 '21 at 05:12
  • By the way, we don’t tend to put $\infty$ at the end of infinite sums like this. – Thomas Andrews May 08 '21 at 05:13
  • I apologize, Mr. Andrews, I actually meant to write "sum of elements of a continuous, finite set of real numbers" :-) – Upayan De May 08 '21 at 05:18
  • And as for the convention of $\infty$ at the end of sums, I must admit I wasn't aware of that. Thanks for letting me know. – Upayan De May 08 '21 at 05:20
  • Maybe a bit simpler: $$\left(\frac13+\frac23\right)+\left(\frac14+\frac34\right)+\left(\frac15+\frac45\right)+\cdots=1+1+1+\cdots=\infty$$ – bof May 08 '21 at 05:21
  • That's nice – Upayan De May 08 '21 at 05:22
  • I think "continuous, finite set" is a contradiction in terms, at any rate if you are using those mathy terms the way mathematicians use them. Maybe you have your own, private meanings for those words, that you'd like to share with us? – Gerry Myerson May 08 '21 at 05:53
  • I actually realised that all such sets are actually infinite. I understand what you mean to say, Mr. Myerson. Careless of me. – Upayan De May 08 '21 at 06:07
  • I apologise. What I'm referring to are subsets of $\mathbb{R}$ such as $[0,2)$ or $(-12,3)$ but not ones like $(-\infty, 6)$ or $(3.5, \infty)$. I hope you understand what I mean. – Upayan De May 08 '21 at 06:11
  • Those are called "finite intervals", but a "finite set" is a set that contains a finite number of points. – bof May 08 '21 at 06:55
  • If you want to be sure I see a comment intended for me, Upayan, you have to put @Gerry in it. That way, I get notified. – Gerry Myerson May 09 '21 at 00:58

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I do not think that your summation is correct. Your summation consists only of non-recurring rational numbers, so clearly many terms are being missed, i.e., all irrational and rational numbers with recurring decimal expansion.

Ritam_Dasgupta
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