For series (a) $\sum_{n=1}^{\infty} 3\left(\frac{1}{5}\right)^{n-1}$, we can use the formula for the sum of a geometric series, $S_{n} = a \frac{1 - r^n}{1 - r}$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. Here, $a = 3$
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