Question

.converge or diverge?  If it converges, to what value does it converge?

Answer 1

Given the series;

$\sum ^{\infty}_{n\mathop=0}3(\frac{1}{5})^{n-1}$

To obtain the sum of the series above and decide if it converges or diverges, we will

$\begin{gathered} \sum ^{\infty}_{n\mathop{=}0}3(\frac{1}{5})^{n-1}=\sum ^{\infty}_{n\mathop{=}0}3(5)^{-(n-1)} \\ =\sum ^{\infty}_{n\mathop{=}0}3(5)^{(1-n)} \\ =\sum ^{\infty}_{n\mathop{=}0}15\times5^{-n} \\ =15\sum ^{\infty}_{n\mathop{=}0}(\frac{1}{5})^n \end{gathered}$

Simplify the resulting geometric series and decide if it converge or diverge

$\sum ^{\infty}_{n\mathop{=}0}(\frac{1}{5})^n\Rightarrow is\text{ an infinite geometric series, with first term a= 1 and common ratio r=}\frac{1}{5}$

Solve for the sum to infinity of the geometric series

$S_{\infty}=\frac{a}{1-r}=\frac{1}{1-\frac{1}{5}}=\frac{1}{\frac{4}{5}}=\frac{5}{4}$

The sum of the series wil be

$15\sum ^{\infty}_{n\mathop{=}0}(\frac{1}{5})^n\Rightarrow15\times\frac{5}{4}=\frac{75}{4}$

Hence,

$\begin{gathered} \sum ^{\infty}_{n\mathop{=}0}3(\frac{1}{5})^{n-1}=\frac{75}{4} \\ \text{The series converges} \end{gathered}$