Limiting pmf as $n \to \infty$ Consider the simple pmf:
$$p_n (x)=\begin{cases} 1\quad x=2+1/n \\ 0\quad \text{elsewhere} \end{cases}$$
Then my book states that $\lim_{n\to \infty} p_n (x)=0$ for all values of $x$. Is that really the case though? Why can we not say that the probability of $x=2$ equals $1$ as $n \to \infty$? Thank you.
 A: 
Why can we not say that the probability of $x=2$ equals 1 as $n\to\infty$? 

For every $n\geq 1$, $p_n(2)=0$. Hence, the sequence of real numbers
$$
  (p_1(2),p_2(2),p_3(2),\dots) = (0,0,0,\dots)
$$
is a constant sequence with limit equal to zero.
A: Just noticed I misread the problem. The proof I sketch below is if $p_n(x) = \left\{ 
  \begin{array}{l l}
    1 & \quad \text{$x=1+1/n$}\\
    0 & \quad \text{otherwise}
  \end{array} \right.$ but a similar proof holds for the OP's question.
Let $p(x)=0$. Pick any value of $x \notin \{1+1/n : n=1,2,3,\ldots\}$. Then $p_n(x)=0$ so $|p_n(x)-p(x)|=0$. Note that $1 \notin \{1+1/n : n=1,2,3,\ldots\}$ so $p_n(1)=p(1)=0$. 
Now pick any $x\in  \{1+1/n : n=1,2,3,\ldots\}$. Let $x=1+1/m$, for $n\geq m+1$, $p_n(1+1/m)=0$. 
It is even further interesting that although the pmf converges to 0, the corresponding CDF does not!
$F_n(x)=P(X_n\leq x) \overset{D}{\longrightarrow} F(x) = \left\{ 
  \begin{array}{l l}
    0 & \quad \text{$x<1$}\\
    1 & \quad \text{$x\geq 1$}
  \end{array} \right.$ at all points of continuity of $x$. Therefore $X_n \overset{D}{\longrightarrow} X$ where $P(X=1)=1$. Even though the density functions converged...the corresponding probability mass functions do not!
