Let $X_n$ be iid random variables such that $E(X_i)=0$ and $E(X_i^2)=\sigma^2$. Prove that $\left(\frac{1}{\sqrt n} \sum_{k=1}^n X_i\right)_n$ cannot converge in probability.
I'm quite stumped with this problem. Since $\frac{1}{\sqrt n} \sum_{k=1}^n X_i = \sqrt n \left(\frac 1n \sum_{k=1}^n X_i\right)$, $\left(\frac{1}{\sqrt n} \sum_{k=1}^n X_i\right)_n$ converges in distribution to $\mathcal N(0,\sigma^2)$, so the question essentially asks to prove that convergence cannot be improved to convergence in probability.
Supposing for the sake of contradiction that $\left(\frac{1}{\sqrt n} \sum_{k=1}^n X_i\right)_n$ converges to some $Y$ in probability, there exists a subsequence that converges to $Y$ almost surely. I haven't been able to derive anything useful from this...
Any hint is welcome.
Here is a one possible way: Write
$$Z_n = \frac{1}{\sqrt{n}} \sum_{k=1}^{n}X_k$$
and assume that $(Z_n)$ converges in probability to some r.v. $Y$. By the classical CLT we know that $Y \sim \mathcal{N}(0, \sigma^2)$. Then $\sqrt{2}Z_{2n} - Z_n$ converges to $(\sqrt{2}-1)Y$ in probability. On the other hand, we know that
$$\sqrt{2}Z_{2n} - Z_n = \frac{1}{\sqrt{n}} \sum_{k=1}^{n} X_{n+k}$$
has the same distribution as $Z_n$, hence this must converge in distribution to $Y$. This forces that
$$ (\sqrt{2}-1)^2\sigma^2 = \operatorname{Var}((\sqrt{2}-1)Y) = \operatorname{Var}(Y) = \sigma^2, $$
from which we obtain $\sigma = 0$.