# Is $'' \sum_{n = 1}^{\infty} (-1)^n \; \text{is a real number}''$ an invalid statement or a false proposition?

So we're beginning an introductory logic course and my professor is giving examples for valid statements/ propositions - meaningful statements that are either true or false but not both. So he puts forth this one;

$$'' \sum_{n = 1}^{\infty} (-1)^n \; \text{is a real number}''$$

I said it was a false proposition. My argument was the statement claims there is a real number $l$ which is equal to $\sum_{n = 1}^{\infty} (-1)^n$ which is false since there is no real number which is equal to that.

My professor says it was not false since it was not a proposition at all. He said the statement was meaningless saying there was no fathomable meaning to the expression $\sum_{n = 1}^{\infty} (-1)^n$. He said such a thing did not exist.

I countered by saying if such a thing ( $\sum_{n = 1}^{\infty} (-1)^n$ ) did not exist then such a real number cannot also exist and that renders the statement false.

My professor countered by saying "such a real number does not exist" means there is no real number "equal" to $\sum_{n = 1}^{\infty} (-1)^n$. But the equality here cannot be computed/evaluated since one of its arguments is meaningless.

Who is right? And why?

• @whacka: Same argument. Prof. would say its invalid since it cannot be computed on the account you don't have a daughter. I would say you do not have a daughter whose hair is red and hence the proposition is false. Commented Aug 21, 2014 at 7:14
• @whacka: I am not going to believe that until you show me a formal document from the census that proves that you don't have a daughter. Commented Aug 21, 2014 at 7:15
• Ishfaaq, the point is that when we do mathematics we don't limit ourselves to first-order logic. We use set theory to make second, third or higher order statements about our favorite objects ("$\Bbb R$ is order complete" or "$G$ is a non-abelian group with a trivial center", etc.) but internally to $\Bbb R$ you can't quantify over a sequence or over all the natural numbers, you can't formalize limits, and so $\sum(-1)^n$ cannot be expressed in the language. So it isn't a proposition. But in a different context, one can formalize this as a first order property. It's not "just $\Bbb R$" though. Commented Aug 21, 2014 at 7:21
• Doesn't the answer to this question depend on the context? Commented Aug 21, 2014 at 7:22
• @Ishfaaq: Did your professor mention which language you were working with? If it was, say, $L=\{+, . , <\}$ then Asaf is correct (and from his argument it is probably an implicit assumption that he probably should have made explicit). If he had not, then it is a bad example since it can lead to the sort of confusion seen here. Commented Aug 22, 2014 at 1:50

Your professor certainly isn't right that "no fathomable meaning" can be assigned to the expression $\sum_{n=1}^{\infty}(-1)^n$. Otherwise, what is meant by the following statement?

The series $\sum_{n=1}^{\infty}(-1)^n$ is divergent.

But in practice one frequently conflates the description of an infinite series with the limit of its partial sums. Your professor could use this formulation instead:

The sum of the series $\sum_{n=1}^{\infty}(-1)^n$ is a real number.

In this case, the sentence doesn't refer to anything, since the sum doesn't exist. It's not a false statement; it's just nonsensical.

• But formally, if you talk about $\Bbb R$ as a first-order structure, you can't formalize the notion of a limit internally. This is a second-order notion. Yes, it has meaning, but it requires more than first-order can say on $\Bbb R$. Commented Aug 21, 2014 at 7:18
• And he has a point in saying its meaningless. Yes the highlighted sentence makes sense. But the standalone expression $\sum (-1)^n$ has no meaning yes? Commented Aug 21, 2014 at 7:20

First look up the definition of an infinite sum:

$$\sum_{n=0}^{\infty} G(n) = \lim_{k\rightarrow \infty} \sum_{n=0}^{k} G(n)$$

Then look up the declaration of a limit:

$$\forall \Delta_F > 0, \, \exists x_0 ,\, \forall x > x_0 : |F(x) - L| < \Delta_F {\color {red} {\iff \atop \rightarrow}} \lim_{x \rightarrow \infty} G(x) = L$$

And the question comes down to whether you are using $\iff$ or $\rightarrow$ in your declaration of a limit. If your declaration of a limit is uses implication, then your given series is undefined. If your declaration of a limit uses equivalence, then your statement involving your series is false.

Here is a simpler example to illustrate the point. Consider the following axioms:

$$3 \mid x \iff P_1(x) \tag{A}$$ $$3 \mid x \rightarrow P_2(x) \tag{B}$$

Notice the following: $P_1(3)$ is true, $P_1(4)$ is false, $P_2(3)$ is true, and most importantly $P_2(4)$ is undefined. So to answer the question, you have to know what your teacher's definition of a limit is.

To your question of "who is right?": whoever can use their results to solve a problem is right. As long as you don't unsoundly redefine the problem itself, you can use whatever definitions you need to solve a problem.

• It is interesting to see how quickly people disregard the point that you can't formalize limits in first-order logic within $\Bbb R$. Commented Aug 22, 2014 at 3:00
• @AsafKaragila Well I don't see in the post where he specifies that he is working in first order logic, and really, you can't specify much in first order logic anyway until you actually encode a higher order logic as variables and values within first order logic. Induction, partial functions, arbitrary arity functions (as you mentioned) just to name a few things unavailable in FOL. I'd say the main issue is conflating "parsable declaration / proposition" with "defined proposition" all under the umbrella of "well defined", which leads to needless confusion. Commented Aug 22, 2014 at 3:08
• Yes, but in the context of logic, in particular introduction to logic, it is customary to work with first-order logic (and the fact this is from an intro class is written in the first line). Finally, the fact that almost everyone can understand the semantics of "I am live in Isreal" doesn't mean it is a syntactically meaningful sentence. It fails to obey grammar and includes a typo. In logic we care about syntax as well as "Oh, everyone knows what the meaning is..." so we are allowed to be nitpicky. Commented Aug 22, 2014 at 3:13
• (1) Whether a string can be parsed into an abstract syntax tree (2) Whether that abstract syntax tree is a proposition (3) Whether that proposition is decidable by axioms (4) whether those axioms are themselves soundly chosen ... these are all concepts which are all commonly carelessly referred to as "valid" or "well defined", which in late 1800s to the early 1900s was forgivable in the evolving birth of symbolic logic, but nowadays I think it's a bit old fashioned ^_^ But I totally agree with you that we should be nitpicky on a mathforum discussing logic. Commented Aug 22, 2014 at 3:24

I believe you are correct, and your professor is incorrect. The summation is a syntactically valid statement, and it has a canonical semantic meaning. However, a non-existent limit is not a real number in much the same way an unicorn is not a real number. It is certainly valid to ask, and people who know the definition of a real number will simply say no, it is not.

Any good mathematician knows that to decide whether someone is right or wrong you must first define a framework where you can be either right or wrong! So let's say the framework is basic set theory. Whether it's true or false depends on how we phrase things. I would say this: The statement $'' \sum_{n = 1}^{\infty} (-1)^n \; \text{is a real number}''$ is the same as the statement $\text{The x for which } x=\sum_{n = 1}^{\infty} (-1)^n \text{ has }x\in\mathbb{R}\text{''}$.

Or rather,

$\forall_x \left((x=\sum_{n = 1}^{\infty} (-1)^n)\to x\in \mathbb{R}\right)$

(that is read "for all x, if x is equal to the infinite sum, then x is a real number")

This statement is undoubtedly, no arguing, true. Another true statement is $\forall_x(1=2 \to x\in\mathbb{R})$. False implies everything! This is actually a useful feature of mathematical logic. If $A$ is the statement "it is raining" and $B$ is the statement "I have my umbrella", then I might assert "$A\to B$". "If it's raining then I'll have my umbrella." This is equivalent to "$\neg A\vee B$". "I have my umbrella or it's not raining." Now, I'm not a liar ;) so I'll make sure that's the case, but if it never rain, that is, when $A$ is false, that doesn't mean that "$A$ implies $B$" is false! So we say false implies everything.

So, it's undoubtedly true if you phrase things like I did. If you don't, then I can't say.

• What is the domain of discourse on your universal quantifier? It may as well be the real numbers, which would make your statement something like, for all real numbers $x$, if it is equal to something that is certainly not a real number, then it is a real number. This adds nothing; the crux of the question still lies in the definition of $\sum_{n=1}^{\infty}(-1)^n$ Commented Dec 3, 2014 at 18:04
• @Jonny Infinite sums in the area of introductory logic have the well defined meaning $x=\sum_{n=1}^\infty a_n \iff \lim_{k\to\infty}\sum_{n=1}^k a_n=x \iff \forall \varepsilon>0 \exists N \forall k>N \mbox{, } :|(\sum_{n=1}^k a_n)-x|<\varepsilon$. What would be pointless would be getting into alternative definitions of infinite sums, since this is really a question on the technicalities of logic. But it's a good question, and I guess $x$ can be anything with a norm where $x$ plus the real number is defined.
– user18862
Commented Dec 3, 2014 at 18:45
• @Jonny You're right. Come to think of it $\exists_{x\in\mathbb{R}} x=\sum_{n=1}^\infty (-1)^n$ would be more to-the-point and give the opposite answer. But I believe my original view is valid too (if you agree that, whatever set you work in, no such $x$ exists).
– user18862
Commented Dec 3, 2014 at 18:54
• But that's the point! We have to get down to the most fundamental definition to check for validity. I agree, it doesn't much matter what the definition is; the point is that applying the definition results in a valid logical formula. 100% agree with your second comment, I was just typing that! Commented Dec 3, 2014 at 18:55

I'd say that you were correct and that the statement in question is false.

The sequence of finite partial sums of such a series goes [-1, 0, -1, 0, ...]. Thus, if you pick any sufficiently large n for a partial sum, you will get a number which lies infinitely close to a member of the set (-1, 0) (two numbers are infinitely close if the absolute value of their difference is infinitesimal, and 0 is an infinitesimal). Thus, such an infinite sum diverges to the two-member set (-1, 0). That is,

$$\sum_{n = 1}^{\infty} (-1)^n= (-1, 0)$$

(-1, 0) is not a real number, and thus we can tell that the original statement ends up false.