If $X_1,...,X_n$ are independent (no need to be identically distributed), $\sigma_n=Var(\sum_jX_j)$ $$ \sum_{j=1}^n(X_j-EX_j)/\sigma_n \overset{d}{\rightarrow} N(0,1) $$ then prove: $$ n^{-1}\sum_{j=1}^n(X_j-EX_j)\overset{P}{\rightarrow}0 \Leftrightarrow \lim_{n\rightarrow\infty}\frac{\sigma_n}{n}=0 $$
“First Edit”
For the "$\Rightarrow$" part, It seems if "$\color{red}{X_n\overset{P}{\rightarrow}0,Y_n\overset{d}{\rightarrow}N(0,1) \Rightarrow X_n/Y_n\overset{d}{\rightarrow}0}$ " is succeed, then I can naturally deduce that conclusion that $\lim_{n\rightarrow\infty}\frac{\sigma_n}{n}=0$, but I'm not sure if the red part is right.
And for the "$\Leftarrow$" part, $\lim_{n\rightarrow\infty}\frac{\sigma_n}{n}=0$ can deduce that $\frac{\sigma_n}{n}\overset{P}{\rightarrow}0$, so using the slutsky theorem, I have: $$ n^{-1}\sum_{j=1}^n(X_j-EX_j)=\sum_{j=1}^n(X_j-EX_j)/\sigma_n \cdot \frac{\sigma_n}{n}\overset{P}{\rightarrow}0 $$
Am I right about the above contents? Please help me, thanks a lot!
“Second Edit”
With exploring deeper in this question, I find a theroem goes like: $$ \text{If g is a continuous mapping, and if } X_n\overset{d}{\rightarrow}X, \text{then } g(X_n)\overset{d}{\rightarrow} g(X) $$ I'm not sure if I can use this theorem in this problem as follow.
Let $g(x)=1/x$, obviously $g$ is a continuous function. Let $Z_n=\sum_{j=1}^n(X_j-EX_j)/\sigma_n$ and $Z\sim N(0,1)$, then according to to the condition, I have $Z_n\overset{d}{\rightarrow} Z$, so applying the theorem, I get: $$ g(Z_n)=\frac{\sigma_n}{\sum_{j=1}^n(X_j-EX_j)}\overset{d}{\rightarrow}\frac{1}{Z}=g(Z) $$ and then using the Slutsky theorem, $$ \frac{\sigma_n}{n}=g(Z_n)\cdot n^{-1}\sum_{j=1}^n(X_j-EX_j)\overset{d}{\rightarrow} g(Z)\cdot 0 = 0 $$ so $\frac{\sigma_n}{n}\rightarrow 0$.
Can this work? I'm looking forward to your reply, Thanks!
Let $$Z_n^{1}=\sum_{j=1}^n(X_j-EX_j)/\sigma_n$$ and
$$Z_n^{2}=\sum_{j=1}^n(X_j-EX_j)/n$$
For the if part, you have $Z_n^1 \overset{d}{\rightarrow} Z^1$, where $Z^1 \sim N(0,1)$ and $Z_n^2 \overset{p}{\rightarrow} 0$. Here $Z_n^2$ converges in distribution to the constant 0. By Slutsky's theorem $\frac{Z_n^2}{Z_n^1}$ converges in distribution to $\frac{1}{Z^1} \times 0$ i.e. the constant 0, which is equivalent to convergence in probability to 0.