If the sum of series `1+(3)/(x)+(9)/(x^(2))+(27)/(x^(3))+ . . .. " to "oo` is a finite number, then

Options

A. `xlt3`
B. `xgt(1)/(3)`
C. `xlt(1)/(3)`
D. `xgt3`

Answer

To solve the problem, we need to analyze the series given:

The series is:
\[ S = 1 + \frac{3}{x} + \frac{9}{x^2} + \frac{27}{x^3} + \ldots \]

This series is a geometric progression (GP) where:
- The first term \( a = 1 \)
- The common ratio \( r = \frac{3}{x} \)

### Step 1: Identify the conditions for the sum of the GP to be finite
The sum of an infinite geometric series is finite if the absolute value of the common ratio \( r \) is less than 1:
\[ |r| < 1 \]
Thus, we have:
\[ \left| \frac{3}{x} \right| < 1 \]

### Step 2: Solve the inequality
To solve \( \frac{3}{x} < 1 \):
1. Multiply both sides by \( x \) (assuming \( x > 0 \)):
\[ 3 < x \]
This means:
\[ x > 3 \]

### Step 3: Write the formula for the sum of the GP
The sum \( S \) of an infinite GP can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
Substituting the values of \( a \) and \( r \):
\[ S = \frac{1}{1 - \frac{3}{x}} \]

### Step 4: Simplify the expression for the sum
Now simplify the expression:
\[ S = \frac{1}{1 - \frac{3}{x}} = \frac{1}{\frac{x - 3}{x}} = \frac{x}{x - 3} \]

### Step 5: Conclusion
For the sum \( S \) to be finite, we have already established that \( x > 3 \).

### Final Answer:
Thus, the condition for \( x \) is:
\[ x > 3 \]