Find the sum to infinity of the following Geometric Progression:`1,1/3,1/9,...`

Explanation

To find the sum to infinity of the given geometric progression \(1, \frac{1}{3}, \frac{1}{9}, \ldots\), we can follow these steps:

### Step 1: Identify the first term (a)
The first term \(a\) of the geometric progression is:
\[ a = 1 \]

### Step 2: Identify the common ratio (r)
To find the common ratio \(r\), we can divide the second term by the first term:
\[ r = \frac{\text{second term}}{\text{first term}} = \frac{\frac{1}{3}}{1} = \frac{1}{3} \]

### Step 3: Check if the series converges
For a geometric series to converge, the absolute value of the common ratio must be less than 1:
\[ |r| = \left|\frac{1}{3}\right| < 1 \]
Since this condition is satisfied, the series converges.

### Step 4: Use the formula for the sum to infinity
The formula for the sum to infinity \(S\) of a geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Substituting the values of \(a\) and \(r\):
\[ S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} \]

### Step 5: Simplify the expression
To simplify \(\frac{1}{\frac{2}{3}}\), we multiply by the reciprocal:
\[ S = 1 \times \frac{3}{2} = \frac{3}{2} \]

### Conclusion
Thus, the sum to infinity of the given geometric progression is:
\[ \boxed{\frac{3}{2}} \]

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