Sure, here's the solution using MathJax formatting:

The given series is: \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots\)

This is a geometric series with a first term (\(a\)) of 1 and a common ratio (\(r\)) of \(\frac{1}{3}\). The sum to infinity of a geometric series converges only when the absolute value of the common ratio is less than 1. In this case, \(|r| = \left|\frac{1}{3}\right| = \frac{1}{3} < 1\), so the series converges.

The formula for the sum to infinity of a convergent geometric series is:

\text{Sum} = \frac{a}{1 - r}, \)

where \(a\) is the first term and \(r\) is the common ratio.

Plugging in the values:

\(a = 1\),

\(r = \frac{1}{3}\).

Therefore, the sum is:

\text{Sum} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}. \)