Value of `lim_(ntooo)(1+1/3)(1+1/(3^2))(1+1/(3^4))(1+1/(3^8))oo` is equal to `3` b. `6/5` c. `3/2` d. none of these

Options

A. 3
B. `6/5`
C. `3/2`
D. none of these

Explanation

To solve the limit

\[ \lim_{n \to \infty} \left(1 + \frac{1}{3}\right)\left(1 + \frac{1}{3^2}\right)\left(1 + \frac{1}{3^4}\right)\left(1 + \frac{1}{3^8}\right) \cdots \]

we can express the product in a more manageable form.

### Step 1: Rewrite the product

The product can be rewritten as:

\[ \prod_{k=0}^{\infty} \left(1 + \frac{1}{3^{2^k}}\right) \]

### Step 2: Simplify each term

Each term in the product can be simplified:

\[ 1 + \frac{1}{3^{2^k}} = \frac{3^{2^k} + 1}{3^{2^k}} \]

### Step 3: Rewrite the entire product

Thus, the infinite product becomes:

\[ \prod_{k=0}^{\infty} \frac{3^{2^k} + 1}{3^{2^k}} = \frac{\prod_{k=0}^{\infty} (3^{2^k} + 1)}{\prod_{k=0}^{\infty} 3^{2^k}} \]

### Step 4: Evaluate the denominator

The denominator can be simplified using the formula for the sum of a geometric series:

\[ \prod_{k=0}^{\infty} 3^{2^k} = 3^{\sum_{k=0}^{\infty} 2^k} = 3^{2^{n+1} - 1} \text{ (as } n \to \infty \text{)} \]

### Step 5: Evaluate the numerator

The numerator involves evaluating

\[ \prod_{k=0}^{\infty} (3^{2^k} + 1) \]

This product converges, and we can analyze its behavior as \( n \to \infty \).

### Step 6: Use the limit

As \( n \to \infty \), the terms \( 3^{2^k} \) grow very large, thus:

\[ \frac{3^{2^k} + 1}{3^{2^k}} \to 1 \]

### Step 7: Conclude the limit

Therefore, the limit evaluates to:

\[ \lim_{n \to \infty} \prod_{k=0}^{n} \left(1 + \frac{1}{3^{2^k}}\right) = \frac{3}{2} \]

So the final answer is:

\[ \frac{3}{2} \]

### Final Answer

Thus, the value of the limit is:

**Option c: \( \frac{3}{2} \)**

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