The given series is $S = \sum_{n=1}^{\infty} \frac{15}{(-2)^n}$.

We can recognize this as an infinite geometric series of the form:

$S = a + ar + ar^2 + ar^3 + \dots$

Where:

• $a$ is the first term of the series.
• $r$ is the common ratio.

Here, the first term $a = \frac{15}{(-2)^1} = -\frac{15}{2}$, and the common ratio $r = \frac{1}{-2} = -\frac{1}{2}$.

The sum of an infinite geometric series is given by the formula:

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

Substitute $a = -\frac{15}{2}$ and $r = -\frac{1}{2}$ into the formula:

$S = \frac{-\frac{15}{2}}{1 - \left(-\frac{1}{2}\right)} = \frac{-\frac{15}{2}}{1 + \frac{1}{2}} = \frac{-\frac{15}{2}}{\frac{3}{2}}$

Now, simplify:

$S = \frac{-\frac{15}{2}}{\frac{3}{2}} = -\frac{15}{2} \times \frac{2}{3} = -\frac{15}{3} = -5$

Thus, the sum of the series is $\boxed{-5}$.

Would you like further details or have any questions?

Related Questions:

1. How is the common ratio determined in a geometric series?
2. What is the condition for the sum of an infinite geometric series to exist?
3. Can you apply the geometric series sum formula to finite series?
4. What happens if the common ratio $r$ is not between -1 and 1?
5. How does a geometric series differ from an arithmetic series?

Tip:

For an infinite geometric series to have a sum, the absolute value of the common ratio $r$ must be less than 1.