Geometric Series

Evaluate the infinite geometric series ​$0.4 + 0.12 + 0.036 + 0.0108 + \dotsb$

$a_1=0.4\\ q=\dfrac{0.12}{0.40}=0.3\\ \dfrac{lim\ s_n}{n\to \infty}= \dfrac{a_1}{1-q}=\dfrac{0.4}{1-0.3}\\ \color{blue}\dfrac{lim\ s_n}{n\to \infty}=s= \dfrac{4}{7}=0.\overline{571428}$

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