Given data

The series is: \(\frac{9}{100}+\frac{9}{100^2}+\frac{9}{100^3}+...+\frac{9}{100^n}+...\)

Consider the infinite series and compare with it given series,

\(b_1+b_2+b_3+...+b_n+...\)

\(b_n=\frac{9}{100^n}\)

The nth partial sum of the given series is,

\(s_n=\sum_{k=1}^nb_k\)

\(=\sum_{k=1}^n\frac{9}{100^k}\)

The first term of the given series is: \(b=\frac{9}{100}\)

The common ratio is: \(r=\frac{1}{100}\)

The expression for the sum of series is,

\(s_n=\frac{b(1-r^n)}{1-r}\)

Substitute the values in the above expression,

\(s_n=\frac{(\frac{9}{100})(1-(\frac{1}{100})^n)}{1-\frac{1}{100}}\)

\(=\frac{100}{99}(\frac{9}{100}-\frac{9}{100}(\frac{1}{100})^n)\)

\(=\frac{100}{99}(\frac{9}{100}-9(\frac{1}{100^{n+1}}))\)

Thus the formula for nth partial sum of series is \(=\frac{100}{99}(\frac{9}{100}-9(\frac{1}{100^{n+1}}))\)

The expression for the limit of the nth partial sum of the given series is,

\(L=\lim_{n\rightarrow\infty}s_n\)

\(=\lim_{n\rightarrow\infty}(\frac{100}{99}(\frac{9}{100}-9\lim_{n\rightarrow\infty}(\frac{1}{100^{n+1}})))\)

\(=\frac{100}{99}(\frac{9}{100}-9(\frac{1}{100^{\infty}}))\)

\(=\frac{100}{99}(\frac{9}{100}-9(0))\)

\(=\frac{1}{11}\)

The nth partial sum of the given series is \(\frac{1}{11}\), and the sequence of the partial sum converges to limit \(L=\frac{1}{11}\), so the given series converges.

Thus the sum of the series is \(\frac{9}{100}+\frac{9}{100^2}+\frac{9}{100^3}+...+\frac{9}{100^n}+...=\frac{1}{11}\)