if a sequence converges in probability so should its mean ? m i correct? let $\{X_n\}$ be a sequence of random variables 
does having finite limit for  $\operatorname{E}(X_n )$ is equivalent to  convergence of sample mean $\mu_n = (x_1 +x_2 + \cdots+ x_n)/n$ ?
my approach is intuitive : if a sequence converges in probability so should its mean ? m i correct ?
$\lim_{n\to\infty} \operatorname{E}( X_n ) \rightarrow a$, i.e $X_n $ converges in probability, this  implies WLLN holds for $ X_n $, which implies $\mu_n$ sholud also converge in probability 
 A: It seems to me you have this if the $\{X_n\}$ are i.i.d. You can relax this, but need to make some other assumptions in case (see below). In fact, if $ X_1 \cdots X_n$ have mean $k$ and variance $\sigma^2$, $\mu_n = (X_1+\cdots+X_n)/n$ will, in this case, have still mean $k:(k*n/n)$, and variance $\sigma^2/n (\sigma^2*n/n^2)$, obviously converging to $0$. As example of violations of the i.i.d hypothesis:
1) Independence: suppose we are in the extreme case where: $X_1 = X_2 =\cdots = X_n$: of course, in this case,  $\mu_n = (X_1+\cdots+X_n)/n = X_1$, thus its variance would remain $\sigma^2$ for any $n$.
2) Identical distribution: suppose you have $X_i \sim N(0,\sigma^2*n^2)$. Here,  $\mu_n = (X_1+\cdots+X_n)/n$ would have variance $\sigma^2*\frac{\sum_{i=1}^{n}i^2}{n^2} \geq \sigma^2$, thus not converging to $0$. However, if your $X_i$ are independent, have identical mean $k$ and finite variance, you also have convergence: obviously $\mu_n = (X_1+\cdots+X_n)/n$ has mean $k$ again, and, if $C|\sigma^2_n \leq C \forall n$, then its variance will be limited by $C/n^2$, thus converging to $0$. This shows that the i.i.d. of the $X_i$s is a sufficient, ma not necessary, condition.
A: Convergence in probability does not imply convergence of the sample mean to the same value.  To see this, consider a sequence of independent random variables $X_1, X_2, X_3, ...$ that each have distribution $X_k \sim k^p \cdot \text{Bern}(1/k^p)$ for some $0<p<1$.  For this sequence we have:
$$\begin{matrix}
\mathbb{E}(X_k) = 1 & & & \mathbb{V}(X_k) = k^p - 1.
\end{matrix}$$

Theorem: For this sequence we have $\text{plim}_{n \rightarrow \infty} X_n = 0$ and $\text{plim}_{n \rightarrow \infty} \bar{X}_n = 1$.
Proof: We will first show that show that $\text{plim}_{n \rightarrow \infty} X_n = 0$.  For any $0<\epsilon <1$ we have:
$$\lim \limits_{n \rightarrow \infty} \mathbb{P}(|X_n| > \epsilon) = \lim \limits_{n \rightarrow \infty} \mathbb{P}(X_n \geqslant 1) = \lim \limits_{n \rightarrow \infty} \frac{1}{n^p} = 0.$$
We now show that $\text{plim}_{n \rightarrow \infty} \bar{X}_n = 1$.  Using Chebyshev's inequality we have:
$$\begin{equation} \begin{aligned}
\lim \limits_{n \rightarrow \infty} \mathbb{P}(|\bar{X}_n - 1| > \epsilon) &\leqslant \lim \limits_{n \rightarrow \infty} \frac{1}{\epsilon^2} \mathbb{V}(\bar{X}_n) \\
&= \lim \limits_{n \rightarrow \infty} \frac{1}{\epsilon^2} \frac{1}{n^2} \sum_{k=1}^n (k^p-1) \\
&< \lim \limits_{n \rightarrow \infty} \frac{1}{\epsilon^2} \frac{1}{n^2} \sum_{k=1}^n n^p \\
&= \lim \limits_{n \rightarrow \infty} \frac{1}{\epsilon^2} \frac{1}{n^{1-p}} =0.
\end{aligned} \end{equation}$$

This particular example uses a pathological sequence with $\lim _{n \rightarrow \infty} \mathbb{V}(X_k) = \infty$.  For this sequence, there is a constant expected value, but there is convergence in probability to a  different value.  Perhaps the best way to get the desired result would be to impose some additional requirement like finite variance to avoid this problem.
