Normal Variables Almost Sure Convergence Ok so I got a question and I think my idea is not appropriate since those are not iid (I thought about using the Strong Law of Large Number somehow) , but I can't think about another way to solve it:
$ Let\\ X_n∼ N(7n,n)\\ and\ \\Y_n=X_n/n \\  show \ that \ Y _n ∼ L \ almost \ surely \\What\ is\ L?$
EDIT:
ok I'll specify my problem - I don't understand the almost-sure convergence term thoroughly.
Therefore I don't know what are the ways to proof that a variable converges almost surely.
 A: Well First thing is to guess what might $L$ be. So if you view $X_{n}=\sum_{i=1}^{n}Z_{i}$ where $Z_{i}$'s are iid $N(7,1)$ variates. Then
$\sum_{i=1}^{n}Z_{i}\sim N(7n,n)$.
Thus $\frac{1}{n}\sum_{i=1}^{n}Z_{i}\xrightarrow{a.s}7$.
Hence $X_{n}$ converges in distribution to $7$. Hence if $X_{n}$ tends almost surely to something, it must be $7$.
So Now for fixed $\epsilon$ .
$$P(|\frac{X_{n}}{n}-7|\geq \epsilon)=P(|\frac{X_{n}-7n}{n}|\geq \epsilon)=P(|\frac{X_{n}-7n}{\sqrt{n}}|\geq\epsilon\sqrt{n})=\\\frac{1}{\sqrt{2\pi}}\int_{\epsilon\sqrt{n}}^{\infty}e^{\frac{-x^{2}}{2}}\,dx+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{-\epsilon\sqrt{n}}e^{\frac{-x^{2}}{2}}\,dx$$
Now you have to show that $\sum_{n=1}^{\infty}\frac{1}{\sqrt{2\pi}}\int_{\epsilon\sqrt{n}}^{\infty}e^{\frac{-x^{2}}{2}}\,dx$  converges. and similarly $\sum_{n=1}^{\infty}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{-\epsilon\sqrt{n}}e^{\frac{-x^{2}}{2}}\,dx$ converges.
Then it follows by Borel-Cantelli Lemma that $\frac{X_{n}}{n}\xrightarrow{a.s} 7$.
