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?

 A: 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.
A: 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.
A: 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.
A: 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.
A: 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.    
