My proof that $S_n/\sqrt n$ does not converge in probability I'm given a sequence $(X_n)$ of  i.i.d. random variables with mean $0$ and finite variance $\sigma^2$. Let $S_n=X_1 + ... + X_n$. I have to show that $S_n/\sqrt n$ does not converge in probability. Here's what I did.
Since $S_n/\sqrt n$ converges in distribution to a normal random variable $Z$ with mean zero, if $S_n/\sqrt n$ converges in probability at all it must be to $Z$. But
$$P(|\frac {S_n} {\sqrt n} - Z| > \epsilon) \geq P(|\frac {S_n} {\sqrt n}|<\epsilon, |Z|>2\epsilon)$$
Now I get to the main point I'm not sure of. Can I say that the random variables $S_n\sqrt n$ (for any $n$) and $Z$ are independant? It seems like they might be, since in some sense we can't tell what the limit of a sequence will be from any initial segment of it.
If so, then I can continue (this all seems correct to me)
$$\begin{align}
&P(|\frac {S_n} {\sqrt n}|<\epsilon, |Z|>2\epsilon) \\
=&P(|\frac {S_n} {\sqrt n}|<\epsilon)P(|Z|>2\epsilon) \\
\geq&(1 - \frac {\sigma^2} {n\epsilon^2})P(|Z|>2\epsilon) \to P(|Z|>2\epsilon) > 0
\end{align}$$
 A: One can easily construct an entity to which we can say that "$S_n/\sqrt n$ converges in probability".  
Take for example $W_n = -X_1+X_2+X_3+...+X_n$
Then 
$$\lim_{n\rightarrow \infty} P\left(\left|\frac {S_n}{\sqrt n} - \frac {W_n}{\sqrt n}\right|> \epsilon\right) = \lim_{n\rightarrow \infty} P\left(\left|\frac {2X_1}{\sqrt n} \right|> \epsilon\right) = 0$$
and the criterion for convergence in probability is satisfied.
So I suspect that "$S_n/\sqrt n$ does not converge in probability" must have a more specific and narrow sense in the OP's case.

On another front, the established phrase "$S_n/\sqrt n$ converges in distribution to a random variable Z" sometimes makes us forget that the phenomenon described by "convergence in distribution" is that the sequence of distribution functions $F_n$ of $S_n/\sqrt n$ converges to a certain distribution function $F$. There is really no $Z$ "at the end of the journey" waiting to "become one" with $S_n/\sqrt n$ .  $Z$ is a random variable, a separate entity from the distribution that characterizes it (which characterizes also an infinite number of other such $Z$'s). If there is no random variable, the question "is $S_n/\sqrt n$ independent of $Z$?" cannot even be posed.
A: Let see if this works.
Set $\sigma^2=1$ (if sigma is 0 the statment is true) and assume $V_n=S_n/\sqrt{n}$ converges in probability to a r.v. Z.
By the CLT, Z has standard normal distribution.
Define $W_n=\frac{S_{2n}}{\sqrt{2n}}$. These variables converge to Z in probability too.
Finally, take $T_n=\frac{S_{2n}-S_n}{\sqrt{n}}$. 
T converges in distribution to a standard normal, but in probability to $(\sqrt{2}-1) Z$, because $T_n=\sqrt{2}W_n-V_n$, that has not standard normal distribution.
A: I have a partial answer. I think what you need is the Cauchy criterion for convergence in probability, which says:
The sequence $\left(\frac{S_n}{\sqrt{n}}\right)_{n\geq 1}$ converges in probability if and only if
\begin{align}
P\left(\bigg\vert\frac{S_n}{\sqrt{n}}-\frac{S_m}{\sqrt{m}}\bigg\vert>\epsilon\right)\stackrel{n,m\to\infty}{\longrightarrow} 0\quad \text{for every }\epsilon>0.
\end{align}
However, notice that
\begin{align}
&P\left(\bigg\vert\frac{S_n}{\sqrt{n}}-\frac{S_m}{\sqrt{m}}\bigg\vert>\epsilon\right)=P\left(\frac{S_n}{\sqrt{n}}-\frac{S_m}{\sqrt{m}}>\epsilon\right)+P\left(\frac{S_n}{\sqrt{n}}-\frac{S_m}{\sqrt{m}}<-\epsilon\right)\\
&\geq P\left(\frac{S_n}{\sqrt{n}}>2\epsilon,\frac{S_m}{\sqrt{m}}\le\epsilon\right)+P\left(\frac{S_n}{\sqrt{n}}\le-2\epsilon,\frac{S_m}{\sqrt{m}}>-\epsilon\right)\\
&\geq P\left(\frac{S_n}{\sqrt{n}}>2\epsilon\right)+P\left(\frac{S_n}{\sqrt{n}}\le-2\epsilon\right)+P\left(\frac{S_m}{\sqrt{m}}\le\epsilon\right)+P\left(\frac{S_m}{\sqrt{m}}>-\epsilon\right)-2\\
&\stackrel{n,m\to\infty}{\longrightarrow}2\left(Q(2\epsilon)+Q(-\epsilon)-1\right),
\end{align}
where $Q(x)=\int\limits_{x}^{\infty} \mathcal{N}(0,1)\,dx$.
I do not know how to find an $\epsilon>0$ such that the term inside the brackets is strictly positive. I am not quite sure if we can find one at all. As I said, this is just a partial solution that I have worked out.
Please feel free to suggest any more additions, or strengthening of arguments.
