Find the sum of infinite terms of the series `: (3)/(2.4) + (5)/(1.4.6) + (7)/(2.4.6.8)+ (9)/(2.4.6.8.10)+……`

Answer

To find the sum of the infinite series given by:

\[ S = \frac{3}{2 \cdot 4} + \frac{5}{1 \cdot 4 \cdot 6} + \frac{7}{2 \cdot 4 \cdot 6 \cdot 8} + \frac{9}{2 \cdot 4 \cdot 6 \cdot 8 \cdot 10} + \ldots \]

we can analyze the pattern in the series.

### Step 1: Identify the general term of the series

The numerators of the series are odd numbers starting from 3, which can be expressed as \(2n + 1\) for \(n = 1, 2, 3, \ldots\). The denominators are products of even numbers. The \(n\)-th term can be expressed as:

\[ T_n = \frac{2n + 1}{2 \cdot 4 \cdot 6 \cdots (2n)} \]

### Step 2: Express the denominator

The denominator can be rewritten using factorial notation. The product of the first \(n\) even numbers can be expressed as:

\[ 2^n \cdot n! \]

Thus, the \(n\)-th term becomes:

\[ T_n = \frac{2n + 1}{2^n \cdot n!} \]

### Step 3: Rewrite the series

Now we can write the series as:

\[ S = \sum_{n=1}^{\infty} \frac{2n + 1}{2^n \cdot n!} \]

### Step 4: Split the series

We can split the series into two parts:

\[ S = \sum_{n=1}^{\infty} \frac{2n}{2^n \cdot n!} + \sum_{n=1}^{\infty} \frac{1}{2^n \cdot n!} \]

### Step 5: Evaluate the first sum

The first sum can be simplified:

\[ \sum_{n=1}^{\infty} \frac{2n}{2^n \cdot n!} = 2 \sum_{n=1}^{\infty} \frac{n}{2^n \cdot n!} = 2 \cdot \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n \cdot n!} = 2 \cdot \frac{1}{2} e^{1/2} = e^{1/2} \]

### Step 6: Evaluate the second sum

The second sum is:

\[ \sum_{n=1}^{\infty} \frac{1}{2^n \cdot n!} = e^{1/2} - 1 \]

### Step 7: Combine the results

Now we combine the results:

\[ S = e^{1/2} + (e^{1/2} - 1) = 2e^{1/2} - 1 \]

### Step 8: Final result

Thus, the sum of the infinite series is:

\[ S = 2e^{1/2} - 1 \]