The series given in the image is:

$\sum_{k=0}^{\infty} 7 \left( \frac{4}{9} \right)^k$

This is a geometric series of the form:

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

Where:

• $a = 7$ (the first term of the series)
• $r = \frac{4}{9}$ (the common ratio)

A geometric series converges if the absolute value of the common ratio $|r|$ is less than 1. In this case:

$|r| = \left|\frac{4}{9}\right| = \frac{4}{9} < 1$

Since $|r| < 1$, the series converges. The sum of an infinite geometric series is given by the formula:

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

Substituting the values of $a$ and $r$:

$S = \frac{7}{1 - \frac{4}{9}} = \frac{7}{\frac{9 - 4}{9}} = \frac{7}{\frac{5}{9}} = 7 \times \frac{9}{5} = \frac{63}{5}$

Thus, the sum of the series is:

$S = \frac{63}{5} = 12.6$

Let me know if you need further details!

Here are 5 related questions:

1. How do you determine if a geometric series converges?
2. What happens to the sum formula if $|r| \geq 1$?
3. Can a series with a negative common ratio converge?
4. How would you generalize the sum formula for any geometric series?
5. What is the sum of the series $\sum_{k=0}^{\infty} 7 \left(\frac{2}{3}\right)^k$?

Tip: Always check the value of the common ratio to decide if a geometric series converges or diverges.