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Can we say that

$\sum_{k=1}^{\infty} 1 =s(N)$,

where $s(N)$ denotes the number of natural numbers?

pjs36
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4 Answers4

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The sum written above is divergent... We can't assign it a value. We can, however, say that the set of natural numbers is countably infinite. Additionally, from a set theoretical perspective, we can say that $\Bbb N$ has cardinality (amount of elements) $\aleph_0$; which is defined precisely to be the cardinality of $\Bbb N$. That is, any countably infinite set $S$ has cardinality $\aleph_0$, and additionally, there exists a bijection between $S$ and $\Bbb N$ (which is how we define uncountably infinite sets).

Meow Mix
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Even though this is unorthodoxical and most writers on this page will probably disagree, I see little reason to say no.

It is not appropriate to speak of the number of natural numbers, but the cardinality of $\mathbb N$ is something well defined and denoted by $\aleph_0$.

Then the notation $\sum_{k=1}^\infty$ indeed corresponds to an exhaustive enumeration of the naturals and summing terms $1$ induces a counting of the elements (it generalizes the finite case).

As $\sum_{k=1}^\infty1$ is not defined in the theory of series because it diverges, I guess it is harmless to define it as equivalent to $\aleph_0$, as long as you don't draw incoherent conclusions from it.

Anyway, given that there are no relevant computation rules that pertain (I am not referring to the summation methods for divergent series), I see little use for such a definition.

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Simply put, no, because the sum makes no sense. It should be clear that

$$\sum_{k=1}^N1\stackrel{N \to\infty}\to\infty\implies DNE$$

Thus it merely hints that there is an infinite amount of natural numbers.

One could also, for example, claim that

$$\sum_{k=1}^\infty2>\sum_{k=1}^\infty1$$

the above sums basically translate down to there "there are twice as many whole numbers than even numbers". But on the contrary, this makes no sense, since for every natural number, there exists an even number that corresponds to it. Thus, such reasoning with infinite sums cannot be done this way.

To get an idea on how many natural numbers there are, you would usually refer to set theory, whereupon you will find the cardinality of the naturals to be $\aleph_0$ (more or less a definition though)

Simply Beautiful Art
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    I take some issue with saying that anything here shows there are infinitely many natural numbers -- without knowing that in advance, how would we know the sequence of partial sums diverges? Also it seems a bit odd to say that we find the cardinality of $\Bbb N$ to be $\aleph_0$ when this is more like a definition. – pjs36 Dec 14 '16 at 21:19
  • @pjs36 Yes, and do wait a bit, I'm still typing more. – Simply Beautiful Art Dec 14 '16 at 21:21
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Let $(a_k)$ be a sequence of real numbers. The writing "$\sum_{k=1}^{\infty} a_k$" is a formal writing - it is called the series of general term $a_k$. To a series, one associates the sequence of partial sums $(S_n)$, which is defined as $S_n = \sum_{k=1}^n a_k$. If the sequence $(S_n)$ converges to $S$ (a real number), then we say that the series $\sum_{k=1}^{\infty}a_k$ converges, and we write $\sum_{k=1}^{\infty}a_k = S$. If $(S_n)$ diverges, then we say that $\sum_{k=1}^{\infty}a_k$ diverges; if $S_n \to \pm \infty$, we write $\sum_{k=1}^{\infty}a_k = \pm \infty$, so the $\pm \infty$ here is just a symbol - in this case $\sum_{k=1}^{\infty} a_k$ does not have a "value" or whatever.

There is no such thing as "the number of natural numbers". To get an idea about how we make sense of "how many" natural numbers are there, consult an introductory textbook to set theory.