`lim_(n->oo) {1/1.3+1/3.5+1/5.7+.....+1/((2n+1)(2n+3))` is equal to

Options

A. 1
B. `(1)/(2)`
C. `-(1)/(2)`
D. none of these

Explanation

To solve the limit \( \lim_{n \to \infty} \left( \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2n+1)(2n+3)} \right) \), we can follow these steps:

### Step-by-Step Solution:

1. **Rewrite the Terms**:
Each term in the series can be rewritten as:
\[ \frac{1}{(2k-1)(2k+1)} = \frac{1}{2} \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) \]
for \( k = 1, 2, \ldots, n \).

2. **Substituting the Terms**:
Therefore, the limit can be rewritten as:
\[ \lim_{n \to \infty} \frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \ldots + \left( \frac{1}{2n+1} - \frac{1}{2n+3} \right) \right) \]

3. **Simplifying the Series**:
Notice that this is a telescoping series. Most terms will cancel out:
\[ = \lim_{n \to \infty} \frac{1}{2} \left( 1 - \frac{1}{2n+3} \right) \]

4. **Taking the Limit**:
As \( n \to \infty \), \( \frac{1}{2n+3} \to 0 \):
\[ = \frac{1}{2} \left( 1 - 0 \right) = \frac{1}{2} \]

5. **Final Result**:
Thus, the limit is:
\[ \lim_{n \to \infty} \left( \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2n+1)(2n+3)} \right) = \frac{1}{2} \]