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 can start by analyzing the denominators of the terms.

### Step 1: Identify the pattern in the denominators
The denominators are:
- \( 9 = 3 \times 3 \)
- \( 18 = 3 \times 6 \)
- \( 30 = 3 \times 10 \)
- \( 45 = 3 \times 15 \)
- \( 63 = 3 \times 21 \)

We can see that each denominator can be expressed as \( 3 \times n \), where \( n \) is a sequence of numbers. The sequence of \( n \) values appears to be \( 3, 6, 10, 15, 21, \ldots \).

### Step 2: Identify the sequence of \( n \)
The sequence \( 3, 6, 10, 15, 21, \ldots \) corresponds to the triangular numbers, which can be expressed as:
\[ T_k = \frac{k(k+1)}{2} \]
for \( k = 2, 3, 4, \ldots \).

### Step 3: Rewrite the series
Thus, we can rewrite the series as:
\[ S = \frac{1}{3} \left( \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + \ldots \right) \]

### Step 4: Express the series in terms of triangular numbers
The series inside the parentheses can be expressed as:
\[ \sum_{k=2}^{\infty} \frac{1}{T_k} = \sum_{k=2}^{\infty} \frac{2}{k(k+1)} \]

### Step 5: Simplify the series using partial fractions
We can simplify \( \frac{2}{k(k+1)} \) using partial fractions:
\[ \frac{2}{k(k+1)} = \frac{2}{k} - \frac{2}{k+1} \]

### Step 6: Write the series as a telescoping series
Thus, the series becomes:
\[ \sum_{k=2}^{\infty} \left( \frac{2}{k} - \frac{2}{k+1} \right) \]
This is a telescoping series, where most terms will cancel out.

### Step 7: Evaluate the telescoping series
The sum of the series can be evaluated as:
\[ \left( \frac{2}{2} - \frac{2}{3} \right) + \left( \frac{2}{3} - \frac{2}{4} \right) + \left( \frac{2}{4} - \frac{2}{5} \right) + \ldots \]
This simplifies to:
\[ 2 - \lim_{n \to \infty} \frac{2}{n+1} = 2 \]

### Step 8: Combine with the factor outside the series
Now, we multiply this result by \( \frac{1}{3} \):
\[ S = \frac{1}{3} \times 2 = \frac{2}{3} \]

### Step 9: Final result
Thus, the sum of the infinite series is:
\[ \boxed{\frac{2}{3}} \]