Find the sum of the infinte series `(1)/(9)+(1)/(18)+(1)/(30)+(1)/(45)+(1)/(63)+…`

Options

A. `(1)/(3)`
B. `(1)/(4)`
C. `(1)/(5)`
D. `(2)/(3)`

Explanation

To find the sum of the infinite series \( S = \frac{1}{9} + \frac{1}{18} + \frac{1}{30} + \frac{1}{45} + \frac{1}{63} + \ldots \), we will follow these steps:

### Step 1: Identify the pattern in the denominators
The denominators of the series are \( 9, 18, 30, 45, 63, \ldots \). We can observe that these numbers can be expressed in terms of their factors:
- \( 9 = 3 \times 3 \)
- \( 18 = 3 \times 6 \)
- \( 30 = 3 \times 10 \)
- \( 45 = 3 \times 15 \)
- \( 63 = 3 \times 21 \)

The denominators can be rewritten as \( 3 \times (3, 6, 10, 15, 21, \ldots) \).

### Step 2: Identify the sequence in the parentheses
The sequence \( 3, 6, 10, 15, 21, \ldots \) can be recognized as the triangular numbers, which can be expressed as:
- \( T_n = \frac{n(n+1)}{2} \)

However, we notice that these numbers can also be expressed as:
- \( 3 = \frac{3(3+1)}{2} \)
- \( 6 = \frac{4(4+1)}{2} \)
- \( 10 = \frac{5(5+1)}{2} \)
- \( 15 = \frac{6(6+1)}{2} \)
- \( 21 = \frac{7(7+1)}{2} \)

### Step 3: Rewrite the series
We can express the series as:
\[ S = \frac{1}{3} \left( \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + \ldots \right) \]
This means we can factor out \( \frac{1}{3} \) from the series.

### Step 4: Express the series in a more manageable form
Now, we can express the series as:
\[ S = \frac{1}{3} \sum_{n=1}^{\infty} \frac{1}{T_n} \]
where \( T_n = \frac{n(n+1)}{2} \).

### Step 5: Use the formula for the sum of the series
The series can be simplified using the formula for the sum of the series of the form \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \):
\[ \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) \]
This is a telescoping series, which converges to 1.

### Step 6: Calculate the sum
Thus, we can find:
\[ S = \frac{1}{3} \cdot 1 = \frac{1}{3} \]

### Final Answer
The sum of the infinite series is:
\[ \boxed{\frac{1}{3}} \]