+218 Compute the sum
\(1 + \frac{1}{5} + \frac{2}{5} + \frac{2}{25} + \frac{4}{25} + \frac{4}{125} + \frac{8}{125} + \frac{8}{625} + \dotsb.\)
+239 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 \(\frac{3}{2}\).
