15. The sum of the infinite series

$$\frac{1}{5}-\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+\frac{1}{5^5}-\frac{1}{5^6}+\cdots$$

is equal to

The correct answer is 1/6.

To find the sum of the given infinite series, we first need to identify its type and structure. Let's write down the series:
$$\frac{1}{5}-\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+\cdots$$
This is an infinite geometric series where each term is obtained by multiplying the preceding term by a constant value, known as the common ratio.

Let's identify the key components of the geometric series:
1. The first term, denoted as a, is:
$$a=\frac{1}{5}$$
2. The common ratio, denoted as r, can be found by dividing any term by the preceding term. For instance, dividing the second term by the first term gives:
$$r=\frac{-\frac{1}{5^2}}{\frac{1}{5}}=-\frac{1}{5}$$

The sum S of an infinite geometric series with a common ratio satisfying |r| < 1 is given by the formula:
$$S=\frac{a}{1-r}$$
Since |r| = |-1/5| = 1/5, which is strictly less than 1, the series converges, and we can apply this formula.

Now, let's substitute the values of a and r into the formula:
$$S=\frac{\frac{1}{5}}{1-\left(-\frac{1}{5}\right)}$$
Simplify the denominator:
$$1-\left(-\frac{1}{5}\right)=1+\frac{1}{5}=\frac{6}{5}$$
Now, substitute this back into the sum equation:
$$S=\frac{\frac{1}{5}}{\frac{6}{5}}=\frac{1}{5}\times \frac{5}{6}=\frac{1}{6}$$
Thus, the sum of the infinite series is exactly 1/6.