`log_(e)3-(log_(e)9)/(2^(2))+(log_(e)27)/(3^(2))-(log_(e)81)/(4^(2))+. . . .` is equal to

Options

A. `(log_(e)3)(log_(e)2)`
B. `log_(e)3`
C. `log_(e)2`
D. `(log_(e)5)/(log_(e)3)`

Answer

To solve the expression \( \log_e 3 - \frac{\log_e 9}{2^2} + \frac{\log_e 27}{3^2} - \frac{\log_e 81}{4^2} + \ldots \), we will break it down step by step.

### Step 1: Rewrite the logarithmic terms
We start by rewriting the logarithmic terms in a more manageable form:
\[ \log_e 9 = \log_e (3^2) = 2 \log_e 3 \]
\[ \log_e 27 = \log_e (3^3) = 3 \log_e 3 \]
\[ \log_e 81 = \log_e (3^4) = 4 \log_e 3 \]
Now we can substitute these back into the original expression:
\[ \log_e 3 - \frac{2 \log_e 3}{2^2} + \frac{3 \log_e 3}{3^2} - \frac{4 \log_e 3}{4^2} + \ldots \]

### Step 2: Factor out \( \log_e 3 \)
Next, we can factor out \( \log_e 3 \) from the entire expression:
\[ \log_e 3 \left( 1 - \frac{2}{4} + \frac{3}{9} - \frac{4}{16} + \ldots \right) \]
This simplifies to:
\[ \log_e 3 \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \right) \]

### Step 3: Identify the series
The series inside the parentheses is:
\[ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \]
This is the alternating series for the natural logarithm:
\[ \log(1 + x) \text{ where } x = 1 \]
Thus, we have:
\[ \log(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \]

### Step 4: Combine results
Substituting back, we find:
\[ \log_e 3 \cdot \log_e 2 \]

### Final Answer
Thus, the expression simplifies to:
\[ \log_e 3 \cdot \log_e 2 \]