Find the sum of series `1+1/2+1/3+1/4+1/6+1/8+1/9+1/12+…… oo,` where the term are the reciprocals of the positive integers whose only prime factors are two's and three's :

Answer

To find the sum of the series \( S = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12} + \ldots \), where the terms are the reciprocals of the positive integers whose only prime factors are 2 and 3, we can follow these steps:

### Step 1: Identify the Series
The series consists of the reciprocals of numbers that can be expressed in the form \( 2^m \cdot 3^n \), where \( m \) and \( n \) are non-negative integers. The first few terms of the series are:
- \( 1 = 2^0 \cdot 3^0 \)
- \( \frac{1}{2} = \frac{1}{2^1 \cdot 3^0} \)
- \( \frac{1}{3} = \frac{1}{2^0 \cdot 3^1} \)
- \( \frac{1}{4} = \frac{1}{2^2 \cdot 3^0} \)
- \( \frac{1}{6} = \frac{1}{2^1 \cdot 3^1} \)
- \( \frac{1}{8} = \frac{1}{2^3 \cdot 3^0} \)
- \( \frac{1}{9} = \frac{1}{2^0 \cdot 3^2} \)
- \( \frac{1}{12} = \frac{1}{2^2 \cdot 3^1} \)

### Step 2: Write the Series in a More Manageable Form
We can express the series \( S \) as:
\[ S = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{2^m \cdot 3^n} \]

### Step 3: Factor the Double Sum
We can separate the sums:
\[ S = \left( \sum_{m=0}^{\infty} \frac{1}{2^m} \right) \cdot \left( \sum_{n=0}^{\infty} \frac{1}{3^n} \right) \]

### Step 4: Calculate Each Geometric Series
The first series is a geometric series with first term \( 1 \) and common ratio \( \frac{1}{2} \):
\[ \sum_{m=0}^{\infty} \frac{1}{2^m} = \frac{1}{1 - \frac{1}{2}} = 2 \]

The second series is also a geometric series with first term \( 1 \) and common ratio \( \frac{1}{3} \):
\[ \sum_{n=0}^{\infty} \frac{1}{3^n} = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2} \]

### Step 5: Multiply the Results
Now, we can multiply the results of the two series:
\[ S = 2 \cdot \frac{3}{2} = 3 \]

### Final Answer
Thus, the sum of the series is:
\[ \boxed{3} \]