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I was wondering if it was possible to express infinite divergent sums using set theory (I do apologize if some of the notation is improper)

I wanted to consider $$\sum_{n=1}^{\infty}n$$ i.e. the sum of all natural numbers. I thought you might be able to express this by adding up the cardinality of sets with n elements. So

$card(\{1\}) +card(\{2,3\})+card(\{4,5,6\})+...$ and so on. This then would equal

$card(\{1\} \bigcup\{2,3\}\bigcup\{4,5,6\}\bigcup...)=\mathbb{N}$

Hence $$\sum_{n=1}^{\infty}n=\aleph_0$$

Is my math correct and can this be generalized to compute other infinite sums, even those which involve terms with decimals?

mfarrington
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  • I think that you need ordinals for this, not cardinals. If I understand correctly, $\sum_{n=1}^{\infty} n = \omega$. Or something like that. – Xander Henderson Mar 28 '18 at 20:46
  • Your math is correct, but I can't imagine how it would apply to anything but sums of nonnegative integers, or sums of cardinal numbers, anyway. – saulspatz Mar 28 '18 at 20:48
  • The natural numbers embed into the reals, into the ordinals, and into the cardinals. And while all these have the capability to deal with countably infinite summations (in the reals it's somewhat limited, the other two are fine with any summation actually), the embedding of the natural numbers into these *only* requires that finitary summation (and so on) is respected. So all three have completely different ways of dealing with the infinite sum of all the natural numbers. So the answer is yes, and no, and "well, that's not the point". For more, see my answer to the duplicate question. – Asaf Karagila Mar 28 '18 at 22:42

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Note that using your notation, $$ \sum_{n=1}^\infty 1 = \aleph_0 $$ and the LHS sum is quite simpler than yours. In fact, I do not understand how you can get anything more than $\aleph_0$ on the RHS at all...

gt6989b
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