Would someone be kind enough to correct my error here?


Now starting with the 1st prime number, regroup the sum:

$S=(1+3+5+7+\cdots) + (2+4+6+\cdots)$

$S=(1+3+5+7+\cdots) + 2(1+2+3+\cdots)$

$S=(1+3+5+7+\cdots) + 2S$

Now repeat the same procedure for every prime to acquire:




Does this logic make sense at all?

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    $\begingroup$ "Does this logic make sense at all?" No. You can not just equate sets with the sum of their elements. $\endgroup$ – J.R. Jan 27 '14 at 14:19
  • $\begingroup$ Could you elaborate, please? Should I have written $S$ instead of $\mathbb{N}$ and called it a series? $\endgroup$ – EntangledLoops Jan 27 '14 at 14:20
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    $\begingroup$ That would be a first step. But even then you are talking in clouds since your series diverges. $\endgroup$ – J.R. Jan 27 '14 at 14:21
  • $\begingroup$ Okay, so you're essentially saying they can't be the same because their partial sums are not equivalent? $\endgroup$ – EntangledLoops Jan 27 '14 at 14:22
  • $\begingroup$ You are writing (a) equalities between a set and a series of its elements, which does not make sense; and (b) series (infinite sums) that do not converge — that is, quantities which are not even defined. $\endgroup$ – Clement C. Jan 27 '14 at 14:23

Apart of some convergence problems and incorrect algebra in the very end, you have almost discovered a very rich and powerful identity, the Euler product of the Riemann zeta function. The main problem with your expression is that it diverges, but we can fix this problem by adding a parameter $s\gt 1$: $$1^{-s}+2^{-s}+3^{-s}+\cdots=\frac{1}{1-2^{-s}}\times\frac{1}{1-3^{-s}}\times\frac{1}{1-5^{-s}}\cdots,$$ the product running over the prime numbers.

This is more familiarly rewritten as $$\zeta(s)=\sum_{k\geq 1}k^{-s}=\prod_{p\,\text{prime}}\frac{1}{1-p^{-s}}.$$ I'll add more on this in a better time, but this identity and the zeta function itself are central for analytic number theory.

Your comparison sounds quite beautiful, by the way: if the sum of all numbers is $S$, the sum of all primes is $1-\dfrac{1}{S}$. As written, this is mere playword; nevertheless, in the Euler product is hidden some deep number theory.

  • $\begingroup$ Thanks for actually listening to the idea, rather than focusing on syntactical errors and inexperience. $\endgroup$ – EntangledLoops Aug 2 '19 at 17:20

"Does this logic make sense at all?" No. You can not just equate sets with the sum of their elements, nor can you divide by $\mathbb{N}$. But let us ignore that for a while and assume you had named the sum $S$.

The problem is that your sum $1+2+3+4+\cdots$ diverges and thus manipulations like reordering the terms, summing in a different order etc. are completely unjustified.

You can not just reorder the terms in divergent series, otherwise you end up with nonsense like the following: $$1=1+(1-1)+(1-1)+(1-1)+\cdots=(1+1)+(-1+1)+(-1+1)+\cdots=2$$

But ok, let us put that aside too. Then in your first step you claim

$$1+2+3+4+\cdots=(1+3+5+7+11+\cdots)+2\cdot (1+2+3+\cdots)$$

where in the first sum there are supposed to be all primes. Then where is the number $9$ for example? And $15$? You forgot all odd numbers which are not primes.

"Formally" the equality is of course correct, since both sides equal $\infty$.

  • $\begingroup$ 9 is taken care of because I said "repeat for all primes" which includes 3. So after 3 you have: $S=(1+5+\cdots)+2S+3S$ $\endgroup$ – EntangledLoops Jan 27 '14 at 14:32
  • $\begingroup$ @snd Aha so you mean $+3\cdot(1+2+3+\cdots)+5\cdot(1+\cdots)$ etc. $\endgroup$ – J.R. Jan 27 '14 at 14:33
  • $\begingroup$ Yes, exactly. Does that at least make sense? $\endgroup$ – EntangledLoops Jan 27 '14 at 14:33
  • $\begingroup$ @snd No not at all. Sorry. But what does make sense is the following: instead of $1+2+3+\cdots$ consider the (formal) power series $x+2x^2+3x^3+\cdots$ and then do you manipulations with that. If you plug in numbers $|x|<1$ then it even converges. $\endgroup$ – J.R. Jan 27 '14 at 14:34
  • $\begingroup$ To be clear, the reason I can't factor is that my idea to rearrange is plagued from the outset? In other words, you're saying: $1+2+3+4+\cdots\ne(1+3+5+\cdots)+(2+4+6+\cdots)$ $\endgroup$ – EntangledLoops Jan 27 '14 at 14:35

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