Given the series;
![\sum ^(\infty)_(n\mathop=0)3((1)/(5))^(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/9vgihzao3juor92vlfrxhrz010khgrvv16.png)
To obtain the sum of the series above and decide if it converges or diverges, we will
![\begin{gathered} \sum ^(\infty)_{n\mathop{=}0}3((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*5^(-n) \\ =15\sum ^(\infty)_{n\mathop{=}0}((1)/(5))^n \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jgc9tl6rqwzhu7hex974ni7qfv32984a50.png)
Simplify the resulting geometric series and decide if it converge or diverge
![\sum ^(\infty)_{n\mathop{=}0}((1)/(5))^n\Rightarrow is\text{ an infinite geometric series, with first term a= 1 and common ratio r=}(1)/(5)](https://img.qammunity.org/2023/formulas/mathematics/college/2bogn2l3fj4ahj6yacd4ez4ebk9eazvxw9.png)
Solve for the sum to infinity of the geometric series
![S_(\infty)=(a)/(1-r)=(1)/(1-(1)/(5))=(1)/((4)/(5))=(5)/(4)](https://img.qammunity.org/2023/formulas/mathematics/college/a2zciqxf4pjhwj4v0l4p0oueqh8bmqp730.png)
The sum of the series wil be
![15\sum ^(\infty)_{n\mathop{=}0}((1)/(5))^n\Rightarrow15*(5)/(4)=(75)/(4)](https://img.qammunity.org/2023/formulas/mathematics/college/yje7gtpjh0znk9ioyii9fc35dujbzjd3si.png)
Hence,
![\begin{gathered} \sum ^(\infty)_{n\mathop{=}0}3((1)/(5))^(n-1)=(75)/(4) \\ \text{The series converges} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/uk44af3fec97a47z5cyw5h0k3mvzoi32du.png)