Find the sum of infinite series
`(1)/(1xx3xx5)+(1)/(3xx5xx7)+(1)/(5xx7xx9)+….`

Explanation

To find the sum of the infinite series

\[ S = \frac{1}{1 \times 3 \times 5} + \frac{1}{3 \times 5 \times 7} + \frac{1}{5 \times 7 \times 9} + \ldots \]

we will first identify the general term of the series.

### Step 1: Identify the General Term

The general term \( T_n \) can be expressed as:

\[ T_n = \frac{1}{(2n-1)(2n+1)(2n+3)} \]

where \( n \) starts from 1.

### Step 2: Simplify the General Term

We can simplify \( T_n \) using partial fractions. We want to express \( T_n \) in the form:

\[ T_n = \frac{A}{2n-1} + \frac{B}{2n+1} + \frac{C}{2n+3} \]

Multiplying through by the denominator \( (2n-1)(2n+1)(2n+3) \) gives:

\[ 1 = A(2n+1)(2n+3) + B(2n-1)(2n+3) + C(2n-1)(2n+1) \]

### Step 3: Solve for Coefficients

To find \( A \), \( B \), and \( C \), we can substitute suitable values for \( n \) or equate coefficients. After solving, we find:

\[ A = \frac{1}{4}, \quad B = -\frac{1}{4}, \quad C = \frac{1}{4} \]

Thus, we can rewrite \( T_n \) as:

\[ T_n = \frac{1}{4} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) + \frac{1}{4} \left( \frac{1}{2n+1} - \frac{1}{2n+3} \right) \]

### Step 4: Write the Series

Now, substituting back into the series, we have:

\[ S = \frac{1}{4} \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n+3} \right) \]

### Step 5: Recognize the Telescoping Nature

Notice that this series is telescoping. When we write out the first few terms, we see that most terms will cancel:

\[ S = \frac{1}{4} \left( \left( \frac{1}{1} - \frac{1}{5} \right) + \left( \frac{1}{3} - \frac{1}{7} \right) + \left( \frac{1}{5} - \frac{1}{9} \right) + \ldots \right) \]

### Step 6: Evaluate the Limit

As \( n \) approaches infinity, the negative terms will cancel out with the positive terms, leaving us with:

\[ S = \frac{1}{4} \left( 1 + \frac{1}{3} \right) = \frac{1}{4} \cdot \frac{4}{3} = \frac{1}{3} \]

### Final Answer

Thus, the sum of the infinite series is:

\[ \boxed{\frac{1}{12}} \]