Find the sum of the series to n terms and to infinity : `(1)/(1.3)+ (1)/(3.5) +(1)/(5.7) +(1)/(7.9)+...`

Answer

To find the sum of the series to n terms and to infinity for the series

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

we can follow these steps:

### Step 1: Identify the General Term
The general term of the series can be expressed as:

\[ t_r = \frac{1}{(2r - 1)(2r + 1)} \]

where \( r \) is the term number starting from 1.

### Step 2: Rewrite the General Term
We can simplify the general term using partial fractions:

\[ t_r = \frac{1}{(2r - 1)(2r + 1)} = \frac{1}{2} \left( \frac{1}{2r - 1} - \frac{1}{2r + 1} \right) \]

### Step 3: Write the Sum of the First n Terms
Now, we can express the sum of the first n terms \( S_n \):

\[ S_n = \sum_{r=1}^{n} t_r = \sum_{r=1}^{n} \frac{1}{2} \left( \frac{1}{2r - 1} - \frac{1}{2r + 1} \right) \]

### Step 4: Simplify the Sum
This sum is telescoping. When we expand it, we have:

\[ S_n = \frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \ldots + \left( \frac{1}{2n - 1} - \frac{1}{2n + 1} \right) \right) \]

Most terms cancel out, leaving us with:

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

### Step 5: Final Expression for \( S_n \)
Thus, we can express \( S_n \) as:

\[ S_n = \frac{1}{2} \left( \frac{2n}{2n + 1} \right) = \frac{n}{2n + 1} \]

### Step 6: Find the Sum to Infinity
To find the sum to infinity \( S_{\infty} \), we take the limit as \( n \) approaches infinity:

\[ S_{\infty} = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{n}{2n + 1} \]

Dividing numerator and denominator by \( n \):

\[ S_{\infty} = \lim_{n \to \infty} \frac{1}{2 + \frac{1}{n}} = \frac{1}{2} \]

### Final Answers
- The sum of the series to n terms is:

\[ S_n = \frac{n}{2n + 1} \]

- The sum of the series to infinity is:

\[ S_{\infty} = \frac{1}{2} \]