Find the sum of infinite terms of the series `: (3)/(2.4) + (5)/(2.4.6) + (7)/(2.4.6.8)+ (9)/(2.4.6.8.10)+……`

Answer

To find the sum of the infinite series given by:

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

we will follow these steps:

### Step 1: Identify the nth term of the series
The numerators of the series are \(3, 5, 7, 9, \ldots\), which can be expressed as \(2n + 1\) where \(n\) starts from 1. The denominators are products of even numbers. The nth 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 expressed as:

\[ 2 \cdot 4 \cdot 6 \cdots (2n) = 2^n \cdot (1 \cdot 2 \cdot 3 \cdots n) = 2^n \cdot n! \]

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

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

### Step 3: Rewrite the series
Now we can write the sum of 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 series
The first series can be simplified:

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

Using the identity \(\sum_{n=0}^{\infty} \frac{n}{n!} x^n = x e^x\), we can evaluate:

\[ \sum_{n=1}^{\infty} \frac{n}{n!} \left(\frac{1}{2}\right)^n = \frac{1}{2} e^{1/2} \]

Thus,

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

### Step 6: Evaluate the second series
The second series is a standard exponential series:

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

### Step 7: Combine the results
Now, we can combine both parts:

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

### Step 8: Final result
Thus, the sum of the infinite series is:

\[ S = 2\sqrt{e} - 1 \]