I got a bit confused during the end of this proof so I am asking for a check. Take $$Y(n) = \begin{cases} 1 &\mbox{with probability} \ 1 -p_n \\ n & \mbox{with probability} \ p_n \end{cases} $$
Assume $p_n \rightarrow 0$ prove that $Y_n$ converges in probability to 1.
My proof:
We want to show that $\lim_{n \rightarrow \infty} P(|Y_n -1| < \epsilon) = 1$, notice that $P(|Y_n -1| < \epsilon) = P(Y_n < \epsilon+1 \ \mbox{or} \ -\epsilon+1 < Y_n \ )$.
In the case $P(-\epsilon+1 < Y_n \ )$ we have that $P( Y_n > -\epsilon+1 \ ) = 1-p_n +p_n= 1 \ \forall{n}$. So here we are done.
In the case $P(Y_n < \epsilon+1)$ we see that $\forall\epsilon \ge 1$ taking $N_1 = \left \lfloor{\epsilon +1}\right \rfloor $ will ensure that $\forall n > N_1 \ P(Y_n < \epsilon+1)= 1-p_n +p_n= 1 $. If $\epsilon$ is less than one then $P(Y_n < \epsilon+1) = 1-p_n$ but we know that there exists an $N_2$ s.t. $\forall n > N_2 \implies p_n < \epsilon_2$ so choosing $N = \max \{ N_1, N_2 \}$ will make $$|P(Y_n < \epsilon+1) -1| < \epsilon_2$$ $\forall n > N.$
Other ways of solving this are welcome, I have a feeling I made it longer than it needed to be.