To find:

The sum of infinite geometric series.

Given:

The geometric series is \(1+\frac14+\frac{1}{16}+\frac{1}{64}+...\)

Concept used:

The sum of infinite term of the geometric series is

\(S_\infty=\frac{a}{1-r}\quad(r<1)\)

Here, a is first term, r is common ratio( less than 1) and \(S_\infty\) is the sum of the infinite series.

Calculation:

The first term of the geometric series is 1.

The common ratio can be obtained by the ratio of second term by first term.

\(\frac{\frac14}{1}=\frac14\)

The common ratio \(\frac14\) is less than 1.

Substitute \(\frac{1}{4}\) for r and 1 for a in equation

\(S_{\infty}=\frac{1}{1-\frac14}\)

\(=\frac{1}{\frac{4-1}{4}}\)

\(=\frac{4}{3}\)

Thus, the sum of infinite geometric series is \(\frac43\)