Let $X_1, X_2,...$ be a sequence of random variables (not necessarily independent or identically distributed)
Give an example of a sequence such that $\sum_{i=1}^n ({X_i-\mu}) \over \sqrt{n}$ converges in distribution to a standard normal $N(0,1)$ but $\lim_{n \to \infty}\sum_{i=1}^n {X_i \over n}$ is not equal $\mu$
I have not found any example of such a sequence. Iwould really appreciate your help or any suggestions
Writing $Z_n=\sum_{i=1}^n(X_i-\mu)/\sqrt n$ we see $\sum_{i=1}^n X_i /n = Z_n/\sqrt n +\mu$, where $X_n=\mu+ \sqrt n Z_n - \sqrt{n-1}Z_{n-1}$.
So any sequence of r.v.s $Z_n$ converging in distribution to $N(0,1)$ such that $Z_n/\sqrt n$ does not converge to $0$ almost surely will provide a counterexample.
Let $Z\sim N(0,1)$ and let $K_n$ be any sequence of $0,1$-valued r.v.s converging to $0$ in probability but not almost surely. It is then easy to check that the sequence $\sqrt nK_n$ also converges to $0$ in probability but not almost surely. So finally take $Z_n=Z+\sqrt n K_n$. By Slutsky's theorem $Z_n$ converges in distribution to $N(0,1)$, but $Z_n/\sqrt n = Z/\sqrt n + K_n $ does not converge almost surely to $0$, even though is does converge to $0$ in probability.