Find the sum of the infinite geometric series `(1+1/3+1/9+1/27+...oo)`.

Options

A. 1
B. `(3)/(2)`
C. 2
D. `(5)/(2)`

Explanation

To find the sum of the infinite geometric series \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots\), we can follow these steps:

### Step 1: Identify the first term and the common ratio
The first term \(A\) of the series is \(1\). The common ratio \(r\) can be found by dividing the second term by the first term:
\[ r = \frac{\frac{1}{3}}{1} = \frac{1}{3} \]

### Step 2: Check if the common ratio is less than 1
For the formula for the sum of an infinite geometric series to be applicable, the common ratio must be less than 1 in absolute value. Here, \(r = \frac{1}{3}\), which is indeed less than 1.

### Step 3: Use the formula for the sum of an infinite geometric series
The sum \(S\) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{A}{1 - r} \]
Substituting the values we have:
\[ S = \frac{1}{1 - \frac{1}{3}} \]

### Step 4: Simplify the expression
Now, simplify the denominator:
\[ 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3} \]
Thus, we can rewrite the sum as:
\[ S = \frac{1}{\frac{2}{3}} = 1 \times \frac{3}{2} = \frac{3}{2} \]

### Conclusion
The sum of the infinite geometric series \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots\) is \(\frac{3}{2}\).

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