To solve for $C$, we need to consider that the total probability must equal 1. This is because the probability distribution $P(X=n)$ sums over all possible values of $n$, and the sum of all probabilities must equal 1.

Given the probability mass function $P(X=n) = C \left(\frac{2}{3}\right)^n$, where $n = 1, 2, 3, \ldots$, we need to sum the series:

$\sum_{n=1}^{\infty} P(X=n) = \sum_{n=1}^{\infty} C \left(\frac{2}{3}\right)^n = 1$

This is a geometric series with the first term $a = C \left(\frac{2}{3}\right)$ and the common ratio $r = \frac{2}{3}$.

The sum of an infinite geometric series is given by:

$\sum_{n=1}^{\infty} ar^n = \frac{a}{1-r}$

Substituting $a = C \left(\frac{2}{3}\right)$ and $r = \frac{2}{3}$ into this formula:

$\sum_{n=1}^{\infty} C \left(\frac{2}{3}\right)^n = C \left(\frac{2}{3}\right) \frac{1}{1-\frac{2}{3}} = C \left(\frac{2}{3}\right) \frac{1}{\frac{1}{3}} = C \left(\frac{2}{3}\right) \times 3 = 2C$

Since the sum must equal 1, we have:

$2C = 1$

Solving for $C$:

$C = \frac{1}{2}$

Thus, the value of $C$ is $\frac{1}{2}$.

Would you like any further details or have any questions? Here are five related questions you might find interesting:

1. What is the expected value $E(X)$ for this probability distribution?
2. How would the solution change if the probability function had a different form?
3. What is the variance $\text{Var}(X)$ for this distribution?
4. How does the sum of an infinite geometric series apply to probability theory in general?
5. How can we verify if a function is a valid probability mass function?

Tip: In probability, the sum of all probabilities in a discrete distribution must always equal 1. This principle is key to solving for constants in probability mass functions.