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Note on the title: This conjecture is not restricted to real numbers, 'decreasing' and 'increasing' were used because of the character limit. The actual conjecture is that every sequence whose terms strictly increase or decrease in magnitude contains at least one term whose magnitude is $0$.

Warning: the following contains a gratuitous excess of '$\infty$' it is recommended that you be familiar with at least one of the following: extended real/complex numbers (projective real, hypereal, Riemann sphere, etc.), non-standard numbers, or "Indian" infinities (Jaina infinities, Bhaskara, etc.).


Let $\mathbb{U}$ denote a universe (e.g. $\mathbb{R}^n,\mathbb{C},\mathbb{H}\setminus\mathbb{R},\ldots$ etc.), where $u\in\mathbb{U}\implies |u|\neq\infty$

Definition: finite - a number* $z\in\mathbb{U}$ is finite iff $|z|\in(0,\infty)$

Definition: infinitesimal - a number $z\in\mathbb{U}$ is infinitesimal iff $\forall w\in\mathbb{U}\setminus\{\mathbb{U}_0\}.|z|<|w|$, where $\mathbb{U_0}$ is the additive identity (zero) of $\mathbb{U}$

Definition: infinite - a number $z\in\mathbb{U}$ is infinite iff $\forall w\in\mathbb{U}.|z|>|w|$

Definition: convergence - the sum of an infinite sequence $A\subseteq \mathbb{U}$ converges iff $\sum A\in\mathbb{U}$


Let $A$ be an infinite sequence in $\mathbb{U}$, indexed by a set $I\subseteq\mathbb{N}\cup\{\infty\}$ where: $$A=\bigcup_{i\in I}\{a_i\}:\qquad\forall i,j\in I.i<j\implies(|a_i|\leq|a_j|\oplus |a_i|\geq|a_j|)$$ and where $\lim_{i\to n}a_i=a_n$, i.e. $A$ is equal to its own limit at every point and is strictly increasing xor decreasing in magnitude.

Conjecture:

$$\sum_{i\in I}a_i\in\mathbb{U}\implies\exists k\in I:|a_k|=0$$

Reasoning:

The sum of infinitely many finite numbers is always infinite (proof needed). Thus, if $A$ contains only finite terms, then the sum of $A$ does not converge.

If $a_1$ is finite and $A$ is increasing, then either all $a\in A$ are finite, or $|a_\infty|=\infty$. In the case of the former, $\sum A$ is the sum of infinitely many finite terms, and thus infinite. In the case of the latter, $A$ contains at least one infinite term, therefore $|\sum A|=|a_1|+\cdots+\infty=\infty$.

This is intuitively reinforced (though not proven) by treating the sum of $A$ as the integral of a continuous function.**

If $A$ is increasing and the sum of $A$ converges, then the first term must be infinitesimal. Since the magnitude of a number is given by a standard, real number, it follows that $|a_1|=0$.

If $A$ is decreasing and the sum of $A$ converges, then $|a_\infty|=0$. Otherwise, $|a_\infty|$ is finite, making all $a\in A$ finite; if this were the case, then sum of $A$ would be the sum of $A$ would not converge.


While this is at least true for all integer sequences (unless there is a proof that the sum of an infinite sequence of increasing integers converges), I'm not absolutely certain that it holds for all sequences. If it does, the how should I prove this? Can I prove this?

I'm particularly skeptical about what would happen if I applied this to sequences of matrices or to sequenences with multiple infinite terms (e.g. where $|a_1|=|a_\infty|=\infty$ but $|a_1+a_\infty|=0$)

*the term "number" can be replaced with "element" as this can also apply to the elements of any $\mathbb{U}$ equipped with a unary operation $|\cdot|:\mathbb{U}\to\mathbb{R}\cup\{\infty\}$ for which $u\in\mathbb{U}\implies|u|\neq\infty$

**Imagine a function $f:\mathbb{R}\to\mathbb{U}$ such that $n\in\mathbb{N}\implies f(n)=a_n$ and $\forall x,y,z\in\mathbb{R}. x<y<z\implies |f(x)|\leq|f(y)|\leq|f(z)|$. For any such function $\int_0^\infty f(x)\ dx=\infty$, even if $f(\infty)$ is finite (proof needed).

This is actually one way to show that $\int\frac{1}{z}\ dx=\ln{z}$.

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    $\begingroup$ And which is the infinitesimal number in the convergent series $\sum_{n=1}^\infty \frac{1}{n^2}$? $\endgroup$ – N. S. Dec 21 '18 at 17:11
  • $\begingroup$ @N.S. the infinitesimal number is $0$. It is the $\infty^\text{th}$ term in the sequence. (following $\frac{1}{\infty^2}=0$) $\endgroup$ – R. Burton Dec 21 '18 at 17:12
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    $\begingroup$ "The sum of infinitely many finite numbers is infinite." This is false. $$1+\frac12+\frac14+\frac18+\cdots=2$$ $\endgroup$ – saulspatz Dec 21 '18 at 17:25
  • $\begingroup$ You are assuming that $a_\infty=\lim_n a_n$, which is not necessary, see my answer. $\endgroup$ – N. S. Dec 21 '18 at 17:26
  • $\begingroup$ @saulspatz by the definitions given, the $\infty^\text{th}$ term of this summation is infinitesimal, not finite. $\endgroup$ – R. Burton Dec 21 '18 at 17:26
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I think there is a fundamental problem with this post, and that is you have abandoned all of the traditional definitions that we would use to discuss this claim. Not to be condescending, but your conjecture is not even wrong; if we can't agree on the definitions then there is no point in arguing over the veracity of your claim.

Take for example the comment on your post given by N.S.; given the sequence $f:\mathbb{N}\rightarrow\mathbb{R}$ defined by $f(n)=1/n^2$, the series $\sum_n f(n)$ converges and no term is $0$. There is no "infinitesimal" term, and there is no "$\infty^{\text{th}}$" element of the sequence.

EDIT: At the request of the OP, I am going to copy and past a lengthy discussion in the comments below into the answer proper.

To pursue this idea further, I would first abandon trying to consider arbitrary sets which have norms on them, and try to formulate your ideas first on $\mathbb{R}$. You want to consider a very unique object, an "infinite sequence with a final term". There is nothing wrong with this idea, but I don't think it has a given name in most circles of mathematics. Allow me to name them hypersequences for the sake of this comment.

Hypersequences of reals are defined to be functions from the set $\mathbb{N}∪\{ω\}$ (the natural numbers plus a symbol $\omega$ which is larger than all naturals in the usual order) to the set of reals. Now you need to define what you mean by convergent hypersequences, what the "sum" of a hypersequence should be, etc. Basically rebuild calculus from the ground up on these new objects. Then, when you have a full accounting of all of the terms and definitions you want to use, attempt to prove that in this new setting, the sum of a convergent hypersequence whose terms either strictly increase or strictly decrease in norm must contain the element $0$.

Once you have a working theory on $\mathbb{R}$, then you can ask yourself "what about this theory really needed to be about the reals?". Assuming that you didn't use any properties specific to the reals, it should then be really easy to generalize to any set that has a notion of magnitude (which should probably just be the idea of a norm).

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  • $\begingroup$ The only reason the definitions were written this way was to avoid taking up space clarifying exactly which sequences this refers to. If you disregard the existence of an $\infty^\text{th}$ term, it just makes the problem more difficult. The labeling was meant to be pragmatic. $\endgroup$ – R. Burton Dec 21 '18 at 17:31
  • $\begingroup$ @R.Burton It is not about diregarding the existence of an $\infty^{\text{th}}$ term; the definition of a sequence simply does not have such a term. Sequences are functions from $\mathbb{N}$ to some other set; a sequence of reals would be a function from naturals to reals, a sequence of complex numbers would be a function from naturals to complex numbers, etc. And the point is that there is no such natural number as $\infty$. $\endgroup$ – DanielOfJack Dec 21 '18 at 17:33
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    $\begingroup$ @R.Burton I would first abandon trying to consider arbitrary sets which have norms on them, and try to formulate your ideas first on $\mathbb{R}$. You want to consider a very unique object, an "infinite sequence with a final term". There is nothing wrong with this idea, but I don't think it has a given name in most circles of mathematics. Allow me to name them hypersequences for the sake of this comment. Hypersequences of reals are defined to be functions from the set $\mathbb{N}\cup\{\omega\}$ (the natural numbers plus a symbol $\omega$ which is larger than all naturals in the usual order) $\endgroup$ – DanielOfJack Dec 21 '18 at 17:46
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    $\begingroup$ @R.Burton to the set of reals. Now you need to define what you mean by convergent hypersequences, what the "sum" of a hypersequence should be, etc. Basically rebuild calculus from the ground up on these new objects. Then, when you have a full accounting of all of the terms and definitions you want to use, attempt to prove that in this new setting, the sum of a hypersequence whose terms either strictly increase or strictly decrease in norm must contain the element $0$. $\endgroup$ – DanielOfJack Dec 21 '18 at 17:49
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    $\begingroup$ @R.Burton Once you have a working theory on $\mathbb{R}$, then you can ask yourself "what about this theory really needed to be about the reals?". Assuming that you didn't use any properties specific to the reals, it should then be really easy to generalize to any set that has a notion of magnitude (which should probably just be the idea of a norm). $\endgroup$ – DanielOfJack Dec 21 '18 at 17:52
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The answer is no. Here is a counterexample: $$a_n= \frac{1}{n^2} \mbox{ if } n\neq \infty \\ a_\infty=-1$$

In your comment you assumed that $a_\infty= \lim_n a_n$ which is not necesarily true. If you add this extra assumption, then your Conjecture is just the divergence test for series, and it is true in each Banach space.

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  • $\begingroup$ Fair enough, I will edit the question. I still need to prove it, though, and this should apply outside of Banach spaces as well (to some extent). $\endgroup$ – R. Burton Dec 21 '18 at 17:40

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