Given decimal: \(0.\overline{81}\)

(a) For the repeating decimal 0.81, we can write

\(0.\overline{81}=0.818181...\)

\(=0.81+0.0081+0.000081+0.00000081+...\)

\(=\frac{81}{10^2}+\frac{81}{10^4}+\frac{81}{10^6}+...\)

\(=\sum_{n=0}^{\infty}\frac{81}{10^2}(\frac{1}{10^2})^n\) (b) Given Series: \(=\sum_{n=0}^{\infty}\frac{81}{10^2}(\frac{1}{10^2})^n\)

Given series is a Geometric series with ratio \(r=\frac{1}{10^2}\) and \(a_1=\frac{81}{10^2}\)

So the sum is \(=\frac{a_1}{(1-r)}\)

\(=\frac{\frac{81}{100}}{(1-\frac{1}{100})}\)

\(=\frac{81}{99}\)

\(=\frac{9}{11}\)

Result

\(\frac{9}{11}\)

(a) For the repeating decimal 0.81, we can write

\(0.\overline{81}=0.818181...\)

\(=0.81+0.0081+0.000081+0.00000081+...\)

\(=\frac{81}{10^2}+\frac{81}{10^4}+\frac{81}{10^6}+...\)

\(=\sum_{n=0}^{\infty}\frac{81}{10^2}(\frac{1}{10^2})^n\) (b) Given Series: \(=\sum_{n=0}^{\infty}\frac{81}{10^2}(\frac{1}{10^2})^n\)

Given series is a Geometric series with ratio \(r=\frac{1}{10^2}\) and \(a_1=\frac{81}{10^2}\)

So the sum is \(=\frac{a_1}{(1-r)}\)

\(=\frac{\frac{81}{100}}{(1-\frac{1}{100})}\)

\(=\frac{81}{99}\)

\(=\frac{9}{11}\)

Result

\(\frac{9}{11}\)