Relation of convergence in probability and almost sure convergence Let $(Y_n)$ real valued random variables, such that $Y_n\to c \in \mathbb R$ in distribution. It has been shown that $Y_n \to c$ in probability. I want to prove that $Y_n \to c $ a.s. does not hold!
Therefore consider $Y_n \sim Ber_{1/n}\,$, i.e. $\mathbb P(Y_n=0)=1-\frac1n$ and $\mathbb P(Y_n=1)=\frac1n$. Then $\mathbb E(Y_n)=\frac1n$. I will use Markow's inequality two times.
$Y_n$ converges in probability to zero $$\lim_{n\to\infty}\mathbb P(\vert Y_n - 0\vert \ge \epsilon)\le \lim_{n\to\infty}\frac{\mathbb E(Y_n)}{\epsilon}=\lim_{n\to\infty}=\frac{\frac1n}{\epsilon}=0.$$ 
$Y_n$ does not converge a.s. to zero. I will use Borel Cantelli,i.e.$$\sum_{n=1}^\infty \mathbb P(\vert Y_n-0\vert \ge \epsilon) \le\sum_{n=1}^\infty\frac{\mathbb E(\vert Y_n\vert)}{\epsilon}=\sum_{i=1}^\infty\frac{\frac1n}{\epsilon}=\infty$$
Are we supposed to assume $\epsilon=1$ and is this attempt fine? If yes, Borel Cantelli provides $Y_n$ does not almost surely converges to zero.  Some comments on this attempt are welcomes!

It seems you don't apply Borel-Cantelli correctly. Consider the events {$Y_n=1$} for $n \in \mathbb{N}$. They are independent and
$$\sum_{n=1}^\infty \mathbb{P}(Y_n=1) = \sum_{n=1}^\infty \frac{1}{n} = \infty$$
Thus by Borel-Cantelli, $\limsup_{n \to \infty} Y_n = 1$.
