Find the sum of n terms and sum to infinity of the following series : `(1)/(2cdot4)+(1)/(4cdot6)+(1)/(6cdot8)+…`

Explanation

To find the sum of the series \( S = \frac{1}{2 \cdot 4} + \frac{1}{4 \cdot 6} + \frac{1}{6 \cdot 8} + \ldots \), we will break down the solution into steps.

### Step 1: Identify the general term of the series
The denominators of the terms in the series can be expressed as products of two consecutive even numbers. The first term is \( 2 \cdot 4 \), the second term is \( 4 \cdot 6 \), the third term is \( 6 \cdot 8 \), and so on.

The \( n \)-th term can be expressed as:
\[ T_n = \frac{1}{(2n)(2n + 2)} = \frac{1}{2n(2n + 2)} \]

### Step 2: Simplify the general term
We can simplify the general term \( T_n \):
\[ T_n = \frac{1}{2n(2n + 2)} = \frac{1}{2n \cdot 2(n + 1)} = \frac{1}{4n(n + 1)} \]

### Step 3: Use partial fractions
We can express \( \frac{1}{n(n + 1)} \) using partial fractions:
\[ \frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1} \]
Thus,
\[ T_n = \frac{1}{4} \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]

### Step 4: Write the sum of the first \( n \) terms
The sum of the first \( n \) terms \( S_n \) can be written as:
\[ S_n = \sum_{k=1}^{n} T_k = \frac{1}{4} \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k + 1} \right) \]
This is a telescoping series. When we expand it, we see that most terms cancel out:
\[ S_n = \frac{1}{4} \left( 1 - \frac{1}{n + 1} \right) \]
So,
\[ S_n = \frac{1}{4} \left( \frac{n}{n + 1} \right) \]

### Step 5: Find the sum to infinity
To find the sum to infinity \( S_{\infty} \), we take the limit of \( S_n \) as \( n \) approaches infinity:
\[ S_{\infty} = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{n}{4(n + 1)} = \lim_{n \to \infty} \frac{1}{4} \cdot \frac{n}{n + 1} \]
Dividing numerator and denominator by \( n \):
\[ S_{\infty} = \frac{1}{4} \cdot \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} = \frac{1}{4} \cdot 1 = \frac{1}{4} \]

### Final Answers
- The sum of the first \( n \) terms is:
\[ S_n = \frac{n}{4(n + 1)} \]
- The sum to infinity is:
\[ S_{\infty} = \frac{1}{4} \]