Given series :

\(\sum_{n=0}^\infty(-\frac15)^n\)

The given series is

\(\sum_{n=0}^\infty(-\frac15)^n=1-\frac{1}{5}+\frac{1}{25}-\frac{1}{125}+....\)

The given series is geometric series.

Here, the first term of the series is a=1.

The common difference is given by

\(r=\frac{-\frac15}{1}\)

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

The formula to find the sum of infinite geometric series is given by

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

Substitute the value of "a" and "r" in the above formula, we get

\(S=\frac{1}{1-(-\frac15)}\)

\(S=\frac{1}{1+\frac15}\)

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

Therefore, the sum of series is \(=\frac{5}{6}\)

\(\sum_{n=0}^\infty(-\frac15)^n\)

The given series is

\(\sum_{n=0}^\infty(-\frac15)^n=1-\frac{1}{5}+\frac{1}{25}-\frac{1}{125}+....\)

The given series is geometric series.

Here, the first term of the series is a=1.

The common difference is given by

\(r=\frac{-\frac15}{1}\)

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

The formula to find the sum of infinite geometric series is given by

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

Substitute the value of "a" and "r" in the above formula, we get

\(S=\frac{1}{1-(-\frac15)}\)

\(S=\frac{1}{1+\frac15}\)

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

Therefore, the sum of series is \(=\frac{5}{6}\)