The series presented is:

$\sum_{k=3}^{\infty} \frac{5}{6^k}$

This is an infinite geometric series, where the general term is given by:

$\frac{5}{6^k}$

We can write this in the form of a geometric series as:

$\frac{5}{6^3} + \frac{5}{6^4} + \frac{5}{6^5} + \cdots$

The general form of a geometric series is:

$\sum_{k=0}^{\infty} ar^k = \frac{a}{1 - r}, \quad \text{where } |r| < 1$

Here, $a = \frac{5}{6^3}$ (the first term when $k = 3$), and $r = \frac{1}{6}$ (the common ratio).

To compute the sum, we use the formula for the sum of an infinite geometric series:

$S = \frac{a}{1 - r}$

Substituting the values $a = \frac{5}{6^3} = \frac{5}{216}$ and $r = \frac{1}{6}$:

$S = \frac{\frac{5}{216}}{1 - \frac{1}{6}} = \frac{\frac{5}{216}}{\frac{5}{6}} = \frac{5}{216} \times \frac{6}{5} = \frac{1}{36}$

Thus, the sum of the series is:

$S = \frac{1}{36}$

Would you like a more detailed explanation on any step or further clarifications? Here are some related questions:

1. How is the general form of a geometric series derived?
2. Can this sum be calculated using other methods, such as telescoping series?
3. How would the sum change if the starting index $k$ were different?
4. What conditions must be met for a geometric series to converge?
5. How does this geometric series relate to real-world applications?

Tip: For a geometric series to converge, the common ratio $r$ must satisfy $|r| < 1$.