The given problem asks to find the sum that converges for the series:

$\sum_{k=1}^\infty 4^{-k}.$

Step 1: Identify the Series

The series is geometric, where the general term is:

$a_k = 4^{-k} = \left( \frac{1}{4} \right)^k.$

A geometric series is of the form:

$\sum_{k=0}^\infty ar^k,$

where $a$ is the first term and $r$ is the common ratio.

In this case:

• The series starts at $k = 1$, not $k = 0$.
• The common ratio is $r = \frac{1}{4}$.

We can rewrite the series starting from $k=1$ as:

$\sum_{k=1}^\infty \left( \frac{1}{4} \right)^k.$

Step 2: Formula for Geometric Series

The sum of an infinite geometric series starting at $k = 0$ is:

$S = \frac{a}{1 - r}, \quad \text{for } |r| < 1.$

Since this series starts at $k = 1$, we modify it slightly. To adjust for the missing $k=0$ term, observe:

$\sum_{k=1}^\infty r^k = \sum_{k=0}^\infty r^k - r^0 = \frac{1}{1 - r} - 1.$

Step 3: Plug in Values

Here, $r = \frac{1}{4}$. Substituting into the formula:

1. First calculate the sum of the series starting at $k = 0$:

$\sum_{k=0}^\infty \left( \frac{1}{4} \right)^k = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}.$

2. Subtract the $k = 0$ term (which is $1$):

$\sum_{k=1}^\infty \left( \frac{1}{4} \right)^k = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3}.$

Final Answer:

The sum of the series is:

$\boxed{\frac{1}{3}}$

Do you need further clarification on geometric series or another part of the problem? Let me know!

Related Questions:

1. How do you determine if a geometric series converges?
2. What is the formula for a finite geometric series?
3. How do you find the common ratio in a geometric sequence?
4. How can geometric series be applied in real-life problems?
5. What happens if the common ratio $|r| \geq 1$?

Tip:

Always check the absolute value of the common ratio $r$. For convergence, $|r|$ must be strictly less than $1$.