The given series is:

$\sum_{k=1}^{\infty} \frac{(-7)^k}{8^{k+1}}$

To solve this, let's manipulate the expression to fit the standard form of a geometric series.

Step 1: Factor out constants

Rewrite the series:

$\sum_{k=1}^{\infty} \frac{(-7)^k}{8^{k+1}} = \frac{1}{8} \sum_{k=1}^{\infty} \left(\frac{-7}{8}\right)^k$

Step 2: Identify as a geometric series

The sum of an infinite geometric series with the form $\sum_{k=0}^{\infty} ar^k$ is:

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

Where $|r| < 1$.

For the given series, we start at $k = 1$, so we need to subtract the first term from the sum:

$S = \sum_{k=1}^{\infty} \left(\frac{-7}{8}\right)^k = \frac{\frac{-7}{8}}{1 - \left(\frac{-7}{8}\right)}$

Step 3: Simplify the expression

Now simplify the denominator:

$S = \frac{\frac{-7}{8}}{1 + \frac{7}{8}} = \frac{\frac{-7}{8}}{\frac{15}{8}} = \frac{-7}{15}$

Step 4: Multiply by $\frac{1}{8}$

Now, multiply the sum by the factor $\frac{1}{8}$ from Step 1:

$\text{Final Sum} = \frac{1}{8} \times \frac{-7}{15} = \frac{-7}{120}$

So, the sum of the series is:

$\boxed{\frac{-7}{120}}$

Would you like any further details or clarifications?

Here are 5 related questions:

1. How does one recognize a geometric series?
2. How is the formula for the sum of an infinite geometric series derived?
3. What are the conditions for the convergence of an infinite geometric series?
4. Can this series be expressed in a different form by reindexing?
5. How can you generalize this method for more complicated series?

Tip: Always check that the common ratio $r$ of a geometric series satisfies $|r| < 1$ for convergence.