The convergence of the sample mean of normal random variables $(X_i)_{1\leq i \leq n}$ are independent independent random variables of the same  distribution $N(m,1)$. Find the distribution of the r.v $\bar{X}_n= \frac1n\sum_{i=1}^n X_i$, then prove that $$\bar{X}_n \to^{a.s} m \quad E|\bar{X}_n - m|^2 \to 0$$
Using m.g.f or using this theorem, $\bar{X}_n= \frac1n\sum_{i=1}^n X_i \sim N(\sum_{i=1}^n \frac{m}{n}, \sum_{i=1}^n \frac{1}{n^2})$. For the rest, I know a lemma from real analysis stating that:
For the sequence $(X_i)_{i \in \mathbb{N}} \subseteq \mathbb{R}$, if  $X_n \to 0$, then $\frac{1}{n} \sum_{i=1}^n X_i \xrightarrow[]{n \to \infty} 0.$
 if $X_i \to 0$. 
How can I use the lemma or how can I prove the  second question ?
I want also to say that I am preparing for my exam and I solve questions from previous exams but I am stuck in this one.
 A: This is an odd question, imprecisely matched to the hints and the curriculum of the course it is associated with  The quantity  $\bar X_n$ is, according to the problem statment, already known to have distribution $N(m,1/n)$, and hence converges in distribution to the constant $m$ as $n\to\infty$, and hence in probability to $m$.  
General cultural knowledge tells us that by the strong law of large numbers $\bar X_n\to m$ almost surely, without needing to know the precise distribution of $\bar X_n$.
Assuming one did not know the SLLN but did know that $\bar X_n\sim N(m,1/n)$ and   also the Borel-Cantelli lemma, one could argue that $\sum_{n>1} P(|\bar X_n-m|>1/\log n)<\infty$ after a Chebychev estimate and conclude that $|\bar X_n-m|>1/\log n$ happens only finitely often.
In greater detail.  Since $\bar X_n\sim N(m,m/n)$ we know $$P(|\bar X_n-m|>1/\log n)=P(|\frac Z{\sqrt n}|>\frac1{\log n})=P(|Z|> \frac{\sqrt n}{\log n}),$$ where $Z\sim N(0,1)$.  But $P(|Z|>\lambda)\le E[Z^4]/\lambda^4=3/\lambda^4$ by Chebychev's inequality (or maybe I should call it Markov's). Here we use the fact that the $4$th moment of $Z$ is $3$; that it is
finite is what's really important. So 
$$\sum_n P(|\bar X_n-m|>1/\log n)\le 3\sum_n \frac {(\log n)^4}{ n^2}<\infty.$$
By the Borel-Cantelli lemma this implies that with probability 1, the inequality $|\bar X_n-m|>1/\log n$ holds for only finitely many $n$. This means, with probability $1$ the inquality $|\bar X_n-m|\le 1/\log n$ holds for all $n$ sufficiently large,  which  implies $\bar X_n\to m$.  (One way to remember B-C almost surely. is to remember that a non-negative  rv with a finite expectation is finite with probability one; apply this to the sum of the indicators of the events $E_n$ mentioned in the cited wiki article.)
