# Check my proof regarding convergence in probability

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.

Most of the intermediate work can be skipped. You don't need to consider all possible $\epsilon$--you are always permitted to focus on sufficiently small $\epsilon$. Similarly, you may ignore a finite number of $n$. So assume $0 \lt \epsilon \lt 1$ and restrict the sequence to $n\ge 2$, whence
$$\lim_{n\to\infty}\Pr(|Y_n - 1| \lt \epsilon) =\lim_{n\to\infty}\Pr(Y_n = 1) = \lim_{n\to\infty}1 - p_n = 1,$$
The idea is correct. I think the formula $$\forall n > N_1 \ P(Y_n < \epsilon+1)= 1-p_n +p_n= 1$$ should read $$\forall n > N_1 \ P(Y_n < \epsilon+1)= 1-p_n,$$ because for such $n$'s, $\{Y_n < \epsilon+1\}=\{Y_n=1\}$.
An "other" way could be the following: define $Z_n:=|Y_n-1|$; then $\mathbb P(Z_n=0)=1-p_n$ and $\mathbb P(Z_n=n-1)=p_n$. Fix $\varepsilon\in (0,1)$. Then for $n \geqslant 3$, we have $$\{|Z_n|> \varepsilon \} \subset\{Z_n=n-1\},$$ hence $$\mathbb P\{|Z_n|> \varepsilon \} \leqslant \mathbb P\{Z_n=n-1\}=p_n.$$