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For example, $e=\dfrac{1}{0!}+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\cdots$ has infinite terms when treated with this form, but it is a countable one, because the cardinality of natural numbers is countably infinite, and my question is if a number can have uncountably infinite terms. Thanks.

My thought is that it is impossible because it must be valid to enumerate the terms, which in uncountably infinites you cannot. Is this true? I need some aclarations.

Edit:

I only need non-zero terms, as sugested in the comments, and the number the terms sum to has to be a finite number.

Garmekain
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    If you mean, can a number be the sum of an uncountably infinite number of other numbers, then integrals in calculus are vaguely a good analogy. – J126 Sep 01 '17 at 23:55
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    Let $f(x)=1$ if $x$ is a nonnegative integer and $0$ otherwise. Then $$\sum_{n\in \mathbb R} \frac{f(n)}{n!}=e$$ Has uncountably many terms. Perhaps you mean nonzero terms? – Franklin Pezzuti Dyer Sep 01 '17 at 23:55
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    I don't understand why this question is being poorly received - it's a very reasonable thing to ask. – Noah Schweber Sep 01 '17 at 23:57
  • @NoahSchweber I agree... I've already upvoted it, don't blame me. :P – Franklin Pezzuti Dyer Sep 01 '17 at 23:58
  • All sums of uncountably many positive terms must diverge (that's because at least one of the buckets $(1,\infty), (\frac 12,1],(\frac 13,\frac 12],\cdots$ must contain infinitely many terms). – lulu Sep 01 '17 at 23:59

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Royden, pge 38, problem 20, part a, is to show that, if $E$ is a set of positive (nonzero) numbers, then $$ \sum_{x \in E} x < \infty $$ only if $E$ is countable.

The problem begins: Let $E$ be a set of positive real numbers. We define $\sum_{x \in E} x$ to be $\sup_{F \in \mathcal F} s_F,$ where $\mathcal F$ is the collection of finite subsets of $E$ and $s_F$ is the (finite) sum of the elements of $F.$

Will Jagy
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  • It actually is a great answer to the question. – Garmekain Sep 02 '17 at 00:06
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    What if $E$ is not a set of positive non-zero terms? (The OP states non-zero terms, so they can be negative as well) – Prasun Biswas Sep 02 '17 at 00:08
  • @PrasunBiswas: If E contained negative terms, then using the supplied definition, your sum would be $\sum_{x \in E \cap (0, \infty)} x$ (where summing over the empty set yields $0$), as using negative terms in the sum is strictly non-optimal. Forcing negative terms into the sum is tricky, as it would require additional assumptions to make the sum well-defined. If you force the negative and positive sums to be finite, it's not very interesting. If not, then you run into the problems of conditional convergence of series. Introducing an order would just reduce the problem back to series anyway. – Theo Bendit Sep 02 '17 at 02:06