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 can start by identifying a general term for the series.

### Step 1: Identify the general term
The series can be expressed in terms of a general term. The \( n \)-th term of the series can be written as:
\[ T_n = \frac{1}{(2n)(2n + 2)} = \frac{1}{2n(2n + 2)} \]

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

### Step 3: Use partial fractions
Next, we can express \( T_n \) using partial fractions:
\[ \frac{1}{4n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1} \]
Multiplying through by \( 4n(n + 1) \) gives:
\[ 1 = A(n + 1) + Bn \]
Setting \( n = 0 \):
\[ 1 = A(0 + 1) \implies A = 1 \]
Setting \( n = -1 \):
\[ 1 = B(-1) \implies B = -1 \]
Thus, we have:
\[ \frac{1}{4n(n + 1)} = \frac{1}{4} \left( \frac{1}{n} - \frac{1}{n + 1} \right) \]

### Step 4: Write the sum of n terms
The sum of the first \( n \) terms, \( S_n \), can be expressed as:
\[ S_n = \frac{1}{4} \left( \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k + 1} \right) \right) \]
This is a telescoping series, where most terms cancel out:
\[ S_n = \frac{1}{4} \left( 1 - \frac{1}{n + 1} \right) \]
Thus, we have:
\[ S_n = \frac{1}{4} \left( \frac{n}{n + 1} \right) \]

### Step 5: Find the sum to infinity
To find the sum to infinity, we take the limit as \( n \) approaches infinity:
\[ S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1}{4} \left( \frac{n}{n + 1} \right) = \frac{1}{4} \cdot 1 = \frac{1}{4} \]

### Final Result
Thus, the sum of the first \( n \) terms is:
\[ S_n = \frac{n}{4(n + 1)} \]
And the sum to infinity is:
\[ S = \frac{1}{4} \]