`1 /(1.3.5) + (1)/(3.5.7) + (1)/(5.7.9) +...oo`

Options

A. `(1)/(3)`
B. `(1)/(8)`
C. `3 log 2`
D. `(1)/(12)`

Answer

To solve the series \( S = \frac{1}{1 \cdot 3 \cdot 5} + \frac{1}{3 \cdot 5 \cdot 7} + \frac{1}{5 \cdot 7 \cdot 9} + \ldots \), we can follow these steps:

### Step 1: Rewrite the series
We can express the series in a more manageable form. Notice that each term in the series can be represented as:
\[ S = \sum_{n=0}^{\infty} \frac{1}{(2n+1)(2n+3)(2n+5)} \]

### Step 2: Factor out a constant
To simplify the terms, we can multiply and divide each term by 4:
\[ S = \frac{1}{4} \left( \frac{4}{1 \cdot 3 \cdot 5} + \frac{4}{3 \cdot 5 \cdot 7} + \frac{4}{5 \cdot 7 \cdot 9} + \ldots \right) \]

### Step 3: Rewrite the numerator
Now, we can rewrite the numerator \( 4 \) in terms of the sequence:
\[ 4 = (5 - 1), \quad 4 = (7 - 3), \quad 4 = (9 - 5) \]
Thus, we can express the series as:
\[ S = \frac{1}{4} \left( \sum_{n=0}^{\infty} \left( \frac{5}{(2n+1)(2n+3)(2n+5)} - \frac{1}{(2n+1)(2n+3)(2n+5)} \right) \right) \]

### Step 4: Split the series
This allows us to split the series into two parts:
\[ S = \frac{1}{4} \left( \sum_{n=0}^{\infty} \frac{1}{(2n+1)(2n+3)} - \sum_{n=0}^{\infty} \frac{1}{(2n+3)(2n+5)} \right) \]

### Step 5: Recognize cancellation
Notice that each term cancels with the subsequent term:
\[ S = \frac{1}{4} \left( \frac{1}{1 \cdot 3} - \lim_{n \to \infty} \frac{1}{(2n+3)(2n+5)} \right) \]
As \( n \to \infty \), the limit approaches zero.

### Step 6: Calculate the remaining term
The remaining term is:
\[ \frac{1}{1 \cdot 3} = \frac{1}{3} \]
Thus, we have:
\[ S = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12} \]

### Final Answer
Therefore, the value of the series is:
\[ S = \frac{1}{12} \]