The sum of the series
`(1)/(1.2) + (1.3)/(1.2.3.4) + (1.3.5)/(1.2.3.4.5.6) + ...` to `oo`, is

Options

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

Answer

To find the sum of 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 first identify the general term of the series.

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

The nth term can be observed as follows:

- The numerator consists of the product of the first n odd numbers, which can be expressed as \(1 \cdot 3 \cdot 5 \cdots (2n - 1)\).
- The denominator consists of the product of the first 2n numbers, which can be expressed as \(1 \cdot 2 \cdot 3 \cdots (2n)\).

Thus, the nth term \(T_n\) can be written as:

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

### Step 2: Simplify the nth term

The product of the first n odd numbers can be expressed using the double factorial notation as:

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

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

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

### Step 3: Write the series in summation form

Now, we can express the sum \(S\) as:

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

### Step 4: Recognize the series as part of the exponential function

The series \(\sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x\). If we set \(x = \frac{1}{2}\), we have:

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

### Step 5: Adjust the series to match the exponential function

Notice that our series starts from \(n=1\), so we can write:

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

### Step 6: Final result

Thus, the sum of the series is:

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

### Summary

The sum of the series

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

is

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