Consider the series

\(\sum_{n=0}^\infty\frac{(-1)^n}{5^n}\)

\(\sum_{n=0}^\infty\frac{(-1)^n}{5^n}=\frac{(-1)^0}{5^0}+\frac{(-1)^1}{5^1}+\frac{(-1)^2}{5^2}+\frac{(-1)^3}{5^3}+...\)

\(\sum_{n=0}^\infty\frac{(-1)^n}{5^n}=\frac{1}{1}+\frac{-1}{5}+\frac{1}{5^2}+\frac{-1}{5^3}+...\)

\(\sum_{n=0}^\infty\frac{(-1)^n}{5^n}=1-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...\)

Series sum

\(\sum_{n=0}^\infty\frac{(-1)^n}{5^n}=1-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...\)

It is a infinite geometric series with first term \(a=1\) and common ratio \(r=-\frac{1}{5}\)

Sum of a infinite geometric series is given by

\(S=\frac{a}{1-r}\)

\(S=\frac{1}{1-(-\frac{1}{5})}\)

\(S=\frac{1}{1+\frac{1}{5}}\)

\(S=\frac{1}{\frac{5+1}{5}}\)

\(S=\frac{5}{6}\)

First term is \(=1\)

Sum \(=\frac{5}{6}\)

\(\sum_{n=0}^\infty\frac{(-1)^n}{5^n}\)

\(\sum_{n=0}^\infty\frac{(-1)^n}{5^n}=\frac{(-1)^0}{5^0}+\frac{(-1)^1}{5^1}+\frac{(-1)^2}{5^2}+\frac{(-1)^3}{5^3}+...\)

\(\sum_{n=0}^\infty\frac{(-1)^n}{5^n}=\frac{1}{1}+\frac{-1}{5}+\frac{1}{5^2}+\frac{-1}{5^3}+...\)

\(\sum_{n=0}^\infty\frac{(-1)^n}{5^n}=1-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...\)

Series sum

\(\sum_{n=0}^\infty\frac{(-1)^n}{5^n}=1-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...\)

It is a infinite geometric series with first term \(a=1\) and common ratio \(r=-\frac{1}{5}\)

Sum of a infinite geometric series is given by

\(S=\frac{a}{1-r}\)

\(S=\frac{1}{1-(-\frac{1}{5})}\)

\(S=\frac{1}{1+\frac{1}{5}}\)

\(S=\frac{1}{\frac{5+1}{5}}\)

\(S=\frac{5}{6}\)

First term is \(=1\)

Sum \(=\frac{5}{6}\)