How do I find the constant C? Consider a random experiment with a sample space
$$S=\{1,2,3,⋯\}$$.
Suppose that we know:
$$P(k) = P({k}) = \frac {c}{3^k}$$ for $k=1,2,⋯,$
where c is a constant.
Find c.
Find $P(\{2,4,6\})$.
Find $P(\{3,4,5,⋯\})$
Relevant equations
For any event $A$, $P(A) ≥ 0$.
Probability of the sample space S is $P(S) = 1$.
If $a_1, a_2, a_3$ are disjoint events, then $P(a_1\cup a_2\cup a_3\cup...) = P(a_1) + P(a_2) + P(a_3)...$
Attempt at a solution
If I plug in values for k, as k increases the probability will decrease. 
$$P(k= 1) = \frac c3$$
$$P(k= 2) = \frac c9$$
$$P(k= 3) = \frac {c}{27}$$
However, I am not understanding two main things.
How am I supposed to know when k stops increasing, or does it go to infinity. 
How am I supposed to find the c value without being given any other information? 
Any help is appreciated.
 A: 1) $k$ never stops increasing, it goes to infinity - this is what's implied by the "...".
2) Your set $S$ of elements is infinite, and as you say any probability that you assign to the sample space must give 1 for the whole sample space; $P(S)=1$. These 2 facts imply:
$$ \sum_{k=1}^{+\infty} P(k) = 1. $$
Solving for that condition gives sufficient information to find $c$.
A: You started along a good path. It is true (and important) that
\begin{align}
P(1) &= \frac{c}{3} \\
P(2) &= \frac{c}{9} \\
P(3) &= \frac{c}{27} \\
\end{align}
Since you were given the sample space $S=\{1,2,3,\ldots\}$,
there is no "last" value of $k$.
So the total probability of the sample space is
\begin{align}
 1 = P(S) &= P(1) + P(2) + P(3) + \cdots \\
          &= \frac{c}{3} + \frac{c}{9} + \frac{c}{27} + \cdots \\
\end{align}
This is a non-terminating geometric series that you must evaluate.
There are well-known formulas to do this.
A: Note that by definition of a probability measure, $\mathbb P(\cup^\infty_{n=1} A_n) = \sum^\infty_{n=1}\mathbb P(A_n)$ for pairwise disjoint events events $A_n$. 
The atoms of $S$, i.e. the "basic" events $\{1\}, \{2\}, \{3\}...$ are clearly pairwise disjoint since
$$
\{ k_i\} \cap \{ k_j \} = \emptyset
$$
for $j\neq j$
Also with the fact that $\mathbb P(S)=1$ you have all you need.
