Given the series;
[tex]\sum ^{\infty}_{n\mathop=0}3(\frac{1}{5})^{n-1}[/tex]To obtain the sum of the series above and decide if it converges or diverges, we will
[tex]\begin{gathered} \sum ^{\infty}_{n\mathop{=}0}3(\frac{1}{5})^{n-1}=\sum ^{\infty}_{n\mathop{=}0}3(5)^{-(n-1)} \\ =\sum ^{\infty}_{n\mathop{=}0}3(5)^{(1-n)} \\ =\sum ^{\infty}_{n\mathop{=}0}15\times5^{-n} \\ =15\sum ^{\infty}_{n\mathop{=}0}(\frac{1}{5})^n \end{gathered}[/tex]Simplify the resulting geometric series and decide if it converge or diverge
[tex]\sum ^{\infty}_{n\mathop{=}0}(\frac{1}{5})^n\Rightarrow is\text{ an infinite geometric series, with first term a= 1 and common ratio r=}\frac{1}{5}[/tex]Solve for the sum to infinity of the geometric series
[tex]S_{\infty}=\frac{a}{1-r}=\frac{1}{1-\frac{1}{5}}=\frac{1}{\frac{4}{5}}=\frac{5}{4}[/tex]The sum of the series wil be
[tex]15\sum ^{\infty}_{n\mathop{=}0}(\frac{1}{5})^n\Rightarrow15\times\frac{5}{4}=\frac{75}{4}[/tex]Hence,
[tex]\begin{gathered} \sum ^{\infty}_{n\mathop{=}0}3(\frac{1}{5})^{n-1}=\frac{75}{4} \\ \text{The series converges} \end{gathered}[/tex]