The sum of the series `(1)/(3.5) + (1)/(5.7) + (1)/(7.9)+….` ad infinity is equal to

Options

A. `1//6`
B. `1//3`
C. `1//2`
D. `5//6`

Answer

To find the sum of the series \( S = \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \ldots \) up to infinity, we can follow these steps:

### Step 1: Identify the general term of the series
The general term of the series can be expressed as:
\[ T_n = \frac{1}{(2n + 1)(2n + 3)} \]
where \( n \) starts from 1.

### Step 2: Simplify the general term using partial fractions
We can express \( T_n \) using partial fractions:
\[ T_n = \frac{1}{(2n + 1)(2n + 3)} = \frac{A}{2n + 1} + \frac{B}{2n + 3} \]
Multiplying through by the denominator \((2n + 1)(2n + 3)\) gives:
\[ 1 = A(2n + 3) + B(2n + 1) \]
Expanding this:
\[ 1 = (2A + 2B)n + (3A + B) \]
Setting coefficients equal, we have:
1. \( 2A + 2B = 0 \)
2. \( 3A + B = 1 \)

From the first equation, \( A + B = 0 \) implies \( B = -A \). Substituting into the second equation:
\[ 3A - A = 1 \implies 2A = 1 \implies A = \frac{1}{2}, B = -\frac{1}{2} \]
Thus, we can rewrite \( T_n \):
\[ T_n = \frac{1/2}{2n + 1} - \frac{1/2}{2n + 3} \]

### Step 3: Write the series in summation form
Now, we can express the sum \( S \) as:
\[ S = \sum_{n=1}^{\infty} \left( \frac{1/2}{2n + 1} - \frac{1/2}{2n + 3} \right) \]

### Step 4: Recognize the telescoping nature of the series
This series is telescoping, meaning that many terms will cancel out:
\[ S = \frac{1}{2} \left( \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{9} \right) + \ldots \right) \]
Most terms cancel, and we are left with:
\[ S = \frac{1}{2} \left( \frac{1}{3} - \lim_{n \to \infty} \frac{1}{2n + 3} \right) \]

### Step 5: Evaluate the limit
As \( n \) approaches infinity, \( \frac{1}{2n + 3} \) approaches 0:
\[ S = \frac{1}{2} \left( \frac{1}{3} - 0 \right) = \frac{1}{6} \]

### Final Answer
Thus, the sum of the series is:
\[ \boxed{\frac{1}{6}} \]