What will be the value of `x^(1//2) , x^(1//4) , x^(1//8)` ……. To infinity.

Options

A. `x^2`
B. x
C. `x^(3//2) `
D. `x^3`

Answer

To solve the problem of finding the value of \( x^{1/2}, x^{1/4}, x^{1/8}, \ldots \) to infinity, we can follow these steps:

### Step 1: Identify the series
We have the terms \( x^{1/2}, x^{1/4}, x^{1/8}, \ldots \). We can express this in a more manageable form by rewriting it as:
\[ S = x^{1/2} + x^{1/4} + x^{1/8} + \ldots \]

### Step 2: Rewrite the terms
We can factor out \( x^{1/2} \) from the series:
\[ S = x^{1/2} \left(1 + x^{-1/4} + x^{-3/8} + \ldots \right) \]

### Step 3: Identify the geometric series
The series inside the parentheses can be recognized as a geometric series where:
- The first term \( a = 1 \)
- The common ratio \( r = x^{-1/4} \)

### Step 4: Sum the geometric series
The formula for the sum \( S \) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Applying this formula, we get:
\[ S = \frac{1}{1 - x^{-1/4}} \]

### Step 5: Substitute back into the equation
Now substituting back into our expression for \( S \):
\[ S = x^{1/2} \cdot \frac{1}{1 - x^{-1/4}} \]

### Step 6: Simplify the expression
To simplify further, we rewrite \( 1 - x^{-1/4} \) as:
\[ 1 - x^{-1/4} = \frac{x^{1/4} - 1}{x^{1/4}} \]
Thus, we have:
\[ S = x^{1/2} \cdot \frac{x^{1/4}}{x^{1/4} - 1} \]

### Step 7: Final expression
This gives us:
\[ S = \frac{x^{3/4}}{x^{1/4} - 1} \]

### Conclusion
Thus, the value of the series \( x^{1/2} + x^{1/4} + x^{1/8} + \ldots \) converges to:
\[ \frac{x^{3/4}}{x^{1/4} - 1} \]