Discrete random variable summation problem. $X$ is a discrete random variable taking on the values $X = 1,3,3^2,3^3,\dots,3^m$ and $f(x) = P(X=x)=\dfrac c x$ for a constant $c$. Find $c$.
Solution:
Since $P(X)=1$, we know that $\dfrac c x=1$, so $c=x$. To find x, we have $x = \sum_0^m 3^m$. Since this series summation diverges to infinity, $c = \infty$.
This is a fascinating problem, however, something doesn't seem right ... in other words, how can $x = \infty$? Is the first statement $c=x$ incorrect?

The probability that $X=1$ is $\dfrac{c}{1}$, the probability that $X=3$ is $\dfrac{c}{3}$, and so on. It follows that
$$\frac{c}{1}+\frac{c}{3}+\frac{c}{3^2}+\cdots +\frac{c}{3^m}=1.$$
We want to find the sum of the finite geometric series 
$1+x+x^2+\cdots+x^m$. 
 where $x=1/3$.  Let
$$f(m)=1+x+x^2+\cdots+x^m.$$
 Note that 
$$xf(m)=x+x^2+x^3+\cdots+x^{m+1}.$$
Subtract. We get
$$(1-x)f(m)=1-x^{m+1},$$
and therefore 
$$f(m)=\frac{1-x^{m+1}}{1-x}.$$
Finally, put $x=1/3$ and note that $c=1/f(m)$. 
Remark: If instead of stopping at $3^m$, we continue forever, then instead of the finite sum of the answer, we have an infinite sum. But since $(1/3)^{n+1}$ approaches $0$ as $n\to\infty$, the infinite geometric series has sum $1/(1-x)=3/2$, and therefore we get $c=2/3$.
