`1/(1.2)+(1.3)/(1.2.3.4)+(1.3.5)/(1.2.3.4.5.6)+.....oo`

Options

A. `e-1`
B. `e^(1//2)-1`
C. `e^(1//2)+e`
D. none of these

Answer

To solve the series

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

we will analyze the general term and relate it to the exponential function.

### Step 1: Identify the general term

The first term is \(\frac{1}{1 \cdot 2}\), the second term is \(\frac{1 \cdot 3}{1 \cdot 2 \cdot 3 \cdot 4}\), and the third term is \(\frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}\). We can see that the numerator consists of the product of odd numbers and the denominator consists of the product of even numbers.

The \(n\)-th term can be expressed as:

\[ T_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{1 \cdot 2 \cdot 3 \cdots (2n)} = \frac{(2n)!}{2^n \cdot n!} \]

### Step 2: Rewrite the series using factorials

The series can be rewritten as:

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

This simplifies to:

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

### Step 3: Recognize the series as an exponential function

The series

\[ \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \]

If we set \(x = \frac{1}{2}\), we have:

\[ e^{1/2} = 1 + \frac{1}{2} + \frac{1}{2^2 \cdot 2!} + \frac{1}{2^3 \cdot 3!} + \ldots \]

Thus,

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

### Step 4: Final answer

Therefore, the final result for the series is:

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