To solve the infinite series:

$\sum_{k=0}^{\infty} \frac{7}{8^k}$

we recognize it as a geometric series. The general form of a geometric series is:

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

where:

• $a$ is the first term of the series (when $k = 0$),
• $r$ is the common ratio.

Step 1: Identify $a$ and $r$

In your series:

• The first term $a = \frac{7}{8^0} = 7$.
• The common ratio $r = \frac{1}{8}$ (since each term is multiplied by $\frac{1}{8}$ to get the next term).

Step 2: Apply the Geometric Series Formula

Since $|r| = \frac{1}{8} < 1$, the series converges, and we can use the formula:

$\sum_{k=0}^{\infty} \frac{7}{8^k} = \frac{7}{1 - \frac{1}{8}}$

Step 3: Simplify the Denominator

Calculate $1 - \frac{1}{8}$:

$1 - \frac{1}{8} = \frac{8 - 1}{8} = \frac{7}{8}$

Step 4: Divide

Now, we find:

$\frac{7}{\frac{7}{8}} = 7 \times \frac{8}{7} = 8$

Final Answer

$\sum_{k=0}^{\infty} \frac{7}{8^k} = 8$

Would you like more details, or do you have any questions?

Here are some related questions for further exploration:

1. How would the sum change if the common ratio were $\frac{1}{2}$ instead of $\frac{1}{8}$?
2. What happens to the sum if $k$ starts from $k = 1$ instead of $k = 0$?
3. How can you find the sum if the numerator is variable instead of a constant $7$?
4. What if the common ratio $r$ is greater than 1?
5. How is this solution impacted if the series converges conditionally instead of absolutely?

Tip: Remember that the convergence of a geometric series depends on $|r| < 1$; otherwise, it diverges.