The sum of the series `1 + (1)/(1!) ((1)/(4)) + (1.3)/(2!) ((1)/(4))^(2) + (1.3.5)/(3!) ((1)/(4))^(3)+ ... ` to `infty`, is

Options

A. `sqrt(2)`
B. 2
C. `(1)/(sqrt(2))`
D. none of these

Explanation

To find the sum of the series

\[ S = 1 + \frac{1}{1!} \left(\frac{1}{4}\right) + \frac{1 \cdot 3}{2!} \left(\frac{1}{4}\right)^2 + \frac{1 \cdot 3 \cdot 5}{3!} \left(\frac{1}{4}\right)^3 + \ldots \]

we can recognize that the series resembles the expansion of the binomial theorem. Specifically, we can relate it to the series expansion of \((1-x)^{-n}\).

### Step 1: Identify the general term of the series
The general term of the series can be expressed as:

\[ \frac{(1)(3)(5)\cdots(2n-1)}{n!} \left(\frac{1}{4}\right)^n \]

This can be rewritten using the double factorial notation as:

\[ \frac{(2n)!}{2^n n!} \left(\frac{1}{4}\right)^n = \frac{(2n)!}{(n!)^2} \left(\frac{1}{2}\right)^{2n} \]

### Step 2: Recognize the series as a known function
The series can be recognized as the Taylor series expansion for \((1-x)^{-1/2}\) evaluated at \(x = \frac{1}{4}\):

\[ (1-x)^{-1/2} = \sum_{n=0}^{\infty} \frac{(2n)!}{(n!)^2} \frac{x^n}{4^n} \]

### Step 3: Substitute \(x = \frac{1}{4}\)
Substituting \(x = \frac{1}{4}\) into the function gives us:

\[ (1 - \frac{1}{4})^{-1/2} = \left(\frac{3}{4}\right)^{-1/2} = \frac{1}{\sqrt{\frac{3}{4}}} = \frac{2}{\sqrt{3}} \]

### Step 4: Final result
Thus, the sum of the series is:

\[ S = \frac{2}{\sqrt{3}} \]