Find the value of `1/(log_(3)e) + 1/(log_(3)e^(2)) + 1/(log_(3)e^(4))`+……….. Is:

Options

A. `log_(e)9`
B. 0
C. 1
D. `log_(e) 270`

Answer

To solve the problem \( S = \frac{1}{\log_{3} e} + \frac{1}{\log_{3} e^{2}} + \frac{1}{\log_{3} e^{4}} + \ldots \), we can follow these steps:

### Step 1: Rewrite the terms using the change of base formula
Using the property of logarithms, we have:
\[ \log_{a} b = \frac{1}{\log_{b} a} \]
Thus, we can rewrite each term in the series:
\[ \frac{1}{\log_{3} e} = \log_{e} 3, \quad \frac{1}{\log_{3} e^{2}} = \log_{e} 3^{2} = 2 \log_{e} 3, \quad \frac{1}{\log_{3} e^{4}} = \log_{e} 3^{4} = 4 \log_{e} 3 \]

### Step 2: Express the series in terms of a common factor
Now, we can express the series as:
\[ S = \log_{e} 3 \left( 1 + 2 + 4 + \ldots \right) \]
The series \( 1 + 2 + 4 + \ldots \) is a geometric series with the first term \( a = 1 \) and the common ratio \( r = 2 \).

### Step 3: Find the sum of the geometric series
The sum of an infinite geometric series can be calculated using the formula:
\[ S_{\infty} = \frac{a}{1 - r} \]
For our series:
\[ S_{\infty} = \frac{1}{1 - 2} = \frac{1}{-1} = -1 \]
However, this series diverges because the common ratio \( r = 2 \) is greater than 1. Thus, we need to reconsider the terms.

### Step 4: Correctly analyze the series
Instead, we should consider the series as:
\[ S = \log_{e} 3 \left( 1 + \frac{1}{2} + \frac{1}{4} + \ldots \right) \]
This series is a geometric series with first term \( 1 \) and common ratio \( \frac{1}{2} \).

### Step 5: Calculate the sum of the corrected series
Using the geometric series sum formula:
\[ S_{\infty} = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]

### Step 6: Combine the results
Now substituting back into our expression for \( S \):
\[ S = \log_{e} 3 \cdot 2 = 2 \log_{e} 3 \]

### Step 7: Final expression in terms of logarithm
Using the property of logarithms:
\[ 2 \log_{e} 3 = \log_{e} 3^{2} = \log_{e} 9 \]

Thus, the final value of the expression is:
\[ \boxed{\log_{e} 9} \]