Geometric Series

Let's observe the pattern of the terms in the series:

$$1, \frac{1}{5}, \frac{2}{5}, \frac{2}{25}, \frac{4}{25}, \frac{4}{125}, \frac{8}{125}, \frac{8}{625}, \dotsb$$

We can see that each pair of terms is related to powers of 2 and 5 in the denominator. Specifically, for each $n \geq 1$, the denominator of the $2n$th term is $5^{n}$, and the numerator of the $2n$th term is $2^{n}$. Similarly, the denominator of the $2n + 1$th term is $5^{n}$, and the numerator of the $2n + 1$th term is $2^{n - 1}$.

Let's rewrite the series:

$$1, \frac{1}{5}, \frac{2}{5}, \frac{2}{25}, \frac{4}{25}, \frac{4}{125}, \frac{8}{125}, \frac{8}{625}, \dotsb$$

$$= 1 + \left(\frac{1}{5}\right) + \left(\frac{2}{5}\right) + \left(\frac{2}{25}\right) + \left(\frac{4}{25}\right) + \left(\frac{4}{125}\right) + \left(\frac{8}{125}\right) + \left(\frac{8}{625}\right) + \dotsb$$

$$= 1 + \left(\frac{1}{5}\right) + \left(\frac{2}{5}\right) + \left(\frac{2}{5}\right)\left(\frac{1}{5}\right) + \left(\frac{4}{5}\right)\left(\frac{1}{5}\right) + \left(\frac{4}{5}\right)\left(\frac{1}{5}\right)^{2} + \left(\frac{8}{5}\right)\left(\frac{1}{5}\right)^{2} + \left(\frac{8}{5}\right)\left(\frac{1}{5}\right)^{3} + \dotsb$$

$$= 1 + \left(\frac{1}{5}\right) + \left(\frac{2}{5}\right) + \left(\frac{2}{25}\right) + \left(\frac{4}{25}\right) + \left(\frac{4}{125}\right) + \left(\frac{8}{125}\right) + \left(\frac{8}{625}\right) + \dotsb$$

Now, we can see that each term is a geometric series. The first term is $1$, and the common ratio is $\frac{1}{5}$. So, the sum of this series is:

$$S_1 = \frac{1}{1 - \frac{1}{5}} = \frac{5}{4}$$

For the terms starting from the second one, the common ratio is also $\frac{1}{5}$. So, the sum of this series is:

$$S_2 = \frac{\frac{1}{5}}{1 - \frac{1}{5}} = \frac{\frac{1}{5}}{\frac{4}{5}} = \frac{1}{4}$$

Thus, the total sum of the series is:

$$S = S_1 + S_2 = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2}$$

Therefore, the sum of the given series is $\frac{3}{2}$.