`1/2+1/4 +1/(8(2)!)+1/(16(3)!)+1/(32(4)!)+ .....oo` =

Options

A. e
B. `sqrte`
C. `sqrte/2`
D. `1/e`

Answer

To solve the series \( S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8 \cdot 2!} + \frac{1}{16 \cdot 3!} + \frac{1}{32 \cdot 4!} + \ldots \), we will analyze the terms and relate them to known series expansions.

### Step 1: Identify the pattern in the series
The series can be rewritten as:
\[ S = \sum_{n=0}^{\infty} \frac{1}{2^{n+1} \cdot n!} \]
where the first term corresponds to \( n=0 \) giving \( \frac{1}{2} \), the second term corresponds to \( n=1 \) giving \( \frac{1}{4} \), and so forth.

### Step 2: Factor out the constant
Notice that we can factor out \( \frac{1}{2} \) from the series:
\[ S = \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n \cdot n!} \]

### Step 3: Recognize the series as the exponential function
The series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \) is the Taylor series expansion for \( e^x \). In our case, we have:
\[ \sum_{n=0}^{\infty} \frac{1}{2^n \cdot n!} = e^{1/2} \]

### Step 4: Substitute back into the equation
Now substituting back into our expression for \( S \):
\[ S = \frac{1}{2} e^{1/2} \]

### Step 5: Final result
Thus, we can express the final result as:
\[ S = \frac{e^{1/2}}{2} = \frac{\sqrt{e}}{2} \]

### Conclusion
The value of the series \( S \) is:
\[ \frac{\sqrt{e}}{2} \]