Disproving a misconception about Grandi's Sum I understand most people do not like questions about Grandi's Sum, but I was hoping to ask more about the reasoning flaw behind it and ask if my proof against it makes sense, as I'm taking real analysis this year and was wondering if I'm using the principles I'm learning in the correct fashion.
Essentially, the misconception I'm talking about is the fact that Grandi's sum, which is $1 - 1 + 1 - 1 + 1 - 1 + ... + (-1)^n + ... $, is equivalent to $\frac{1}{2}$. My proof against this claim is as follows, and is actually a proof showing the sum does not equal ANY number at all...
Take the sequence $s_n = \sum\limits_{k=0}^{n} (-1)^k$. That is, $s_0 = 1, s_1 = 1-1 = 0, ...$ and so on. We can construct two subsequences from $s_n$: one of the even-indexed terms of $s_n$, and another of the odd-indexed terms of $s_n$:
$\{s_{2k}\} = 1,1,1 ...$
$\{s_{2k+1}\} = 0, 0, 0 ...$ 
As such, we have:
\begin{align*}
\lim_{k \to \infty} s_{2k} &= 1 \\
\lim_{k \to \infty} s_{2k+1} &= 0 \\
\end{align*}
(By the Subsequence Theorem) If the sequence $s_n$ is convergent, i.e. has a limit $L$, then all of its subsequences are convergent to $L$. However, we have two subsequences of $s_n$ converge to different limits, and so we conclude $s_n$ is not convergent, i.e. its limit does not exist. 
We note the following:
$\lim_{n \to \infty} s_n = \sum\limits_{k=0}^{\infty} (-1)^k$ 
Since the limit of $s_n$ as $n \rightarrow \infty$ does not exist, the summation on the RHS does not converge to a value. 
As such, we cannot propose that $1 - 1 + 1 - 1 + ... + (-1)^n + ...$ is equal to a number $S$, as that implies the summation is convergent, which we have shown is untrue. Thus, the claim $1 - 1 + 1 - 1 + ... + (-1)^n + ... = \frac{1}{2}$ is invalid. 
I think one of the fatal flaws in the claim $1 - 1 + 1 - ... = \frac{1}{2}$ is assuming the convergence of the series in the first place, which I am trying to point out using some of the facts I am currently learning. Is this proof correct? 
 A: You are correct, if you use the standard definition of an infinite sum,[*] there is no limit.
The standard definition of $\sum_{k=1}^{\infty} a_k$ is as: $$\lim_{n\to\infty} \sum_{k=1}^{n} a_k$$
Where $\lim_{n\to\infty}$ is given the usual $\epsilon-N$ definition. 
In this definition, as you say, the series has no limit, and a series cannot converge unless $a_n\to 0.$

There are other ways to talk about infinite sums, but they are non-standard. 
The only ones we really care are ones that agree with the above definition when the above definition exists.
You can think of these definitions like we think of "principal values" of an integral. For example, with the standard definitions:
$$\int_{-1}^{1}\frac{dx}{x}$$ is undefined, but we can define a "principal value" to the integral which, in this case, is zero simply by symmetry.
So we have the notion, for example, of the Cesàro sum of a sequence, or, more complicated, the Ramanujan sum.
Tnere are things that are surprising about these sums. For example, if you add a bunch of zeros between the $1$ and $-1$, the sum is different:
$$1+0+(-1)+1+0+(-1)+\cdots = \frac{2}{3}\quad(\mathfrak C)$$
Where the $\mathfrak C$ is to clarify that we are talking about the Cesàro sum.
What is remarkable is that these various definitions tend to agree, when they both can evaluate the same sum. Also, these sums seem to be of some use. You just have to be very careful manipulating them, because they don't behave as you'd suspect.

[*] And if you are just learning calculus or analysis, you really should only be using that definition for now.
