`3^((1)/(9)).9^((1)/(27)).27^((1)/(81)).81^((1)/(243))….oo=`

Options

A. `sqrt3`
B. 1
C. `(1)/(3)`
D. none of these

Answer

To solve the expression \(3^{\frac{1}{9}} \cdot 9^{\frac{1}{27}} \cdot 27^{\frac{1}{81}} \cdots\) as it goes to infinity, we can rewrite each term with a base of 3.

### Step 1: Rewrite each term
The terms can be rewritten as:
- \(9 = 3^2\)
- \(27 = 3^3\)
- \(81 = 3^4\)

So, we have:
\[ 3^{\frac{1}{9}} \cdot (3^2)^{\frac{1}{27}} \cdot (3^3)^{\frac{1}{81}} \cdots \]

### Step 2: Simplify the powers
Now, we simplify the powers:
\[ 3^{\frac{1}{9}} \cdot 3^{\frac{2}{27}} \cdot 3^{\frac{3}{81}} \cdots \]

### Step 3: Combine the bases
Since the bases are the same, we can combine the exponents:
\[ 3^{\left(\frac{1}{9} + \frac{2}{27} + \frac{3}{81} + \cdots\right)} \]

### Step 4: Find the general term of the exponent
The general term can be expressed as:
\[ \frac{n}{3^{n+1}} \quad \text{for } n = 1, 2, 3, \ldots \]

### Step 5: Sum the series
Now, we need to find the sum of the series:
\[ S = \frac{1}{9} + \frac{2}{27} + \frac{3}{81} + \cdots \]

This can be rewritten as:
\[ S = \sum_{n=1}^{\infty} \frac{n}{3^{n+1}} \]

### Step 6: Use the formula for the sum of an infinite series
Using the formula for the sum of the series \(\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}\) for \(|x| < 1\), we set \(x = \frac{1}{3}\):
\[ \sum_{n=1}^{\infty} n \left(\frac{1}{3}\right)^n = \frac{\frac{1}{3}}{\left(1 - \frac{1}{3}\right)^2} = \frac{\frac{1}{3}}{\left(\frac{2}{3}\right)^2} = \frac{\frac{1}{3}}{\frac{4}{9}} = \frac{3}{4} \]

### Step 7: Adjust for the factor of \(3^{-1}\)
Since we have \(S = \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}\), we can now substitute this back into our exponent:
\[ S = \frac{1}{4} \]

### Step 8: Final expression
Thus, we have:
\[ 3^{\frac{1}{4}} = \sqrt[4]{3} \]

### Conclusion
The final result of the infinite product is:
\[ \sqrt[4]{3} \]