<s>[INST] Valid values of constant for probability mass function (PMF) 
For what values of constant $c$ do the following functions define a valid PMF for random variable $X$ on supports $X=\{1,2,...\}$
(1) $f(x) = c/2^x$
  (2) $f(x) = c2^x/x!$

I was thinking $\sum_{i=1}^\infty f(x) = 1$ so, 
$$\begin{aligned}
\sum_{i=1}^\infty \frac{c}{2^x} &= 1 \\
c &= \sum_{i=1}^\infty 2^x
\end{aligned}$$
and $$\begin{aligned}
\sum_{i=1}^\infty \frac{c2^x}{x!} &= 1 \\
c &= \sum_{i=1}^\infty \frac{x!}{2^x}
\end{aligned}$$
Is these correct? (1) sums to infinity? Not sure about (2)
 [/INST] For the first, your start was correct, we want
$$c\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots\right)=1.$$
The inner sum above is an infinite geometric series, with first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$. 
It has sum is $\dfrac{1/2}{1-1/2}$, which is $1$. 
So we want $(c)(1)=1$, and therefore $c=1$.
For the second problem, again the start was correct, we want
$$c\sum_{n=1}^\infty \frac{2^n}{n!}=1.\tag{1}$$
You may recall that for any $t$, we have 
$$\sum_{n=0}^\infty \frac{t^n}{n!}=e^t.\tag{2}$$
Comparing with the sum in (1), and noting that the first term of (2) is missing in (1), we find that 
$$\sum_{n=1}^\infty \frac{2^n}{n!}=e^2-1.$$
So we want $c(e^2-1)=1$, making $c=\frac{1}{e^2-1}$.  
Remark: What was going wrong in each calculation was the built-in assumption that the reciprocal of a sum (in this case an infinite "sum") is equal to the sum of the reciprocals. To simplify the mistake, you were assuming that the reciprocal $\frac{1}{a+b}$ of $a+b$ is equal to $\frac{1}{a}+\frac{1}{b}$.  This is in general not true. 
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