The value of `1-log_(e)2+(log_(e)2)^(2)/(2!)-(log_(e)2)^(3)/(3!)+..` is

Options

A. 2
B. `1/2`
C. `log_(e)3`
D. none of these

Answer

To solve the expression \( 1 - \log_e 2 + \frac{(\log_e 2)^2}{2!} - \frac{(\log_e 2)^3}{3!} + \ldots \), we can recognize that it resembles the Taylor series expansion of \( e^{-x} \).

### Step-by-Step Solution:

1. **Identify the Series**:
The given series can be rewritten as:
\[ 1 - \log_e 2 + \frac{(\log_e 2)^2}{2!} - \frac{(\log_e 2)^3}{3!} + \ldots \]
This is the Taylor series expansion for \( e^{-x} \) where \( x = \log_e 2 \).

2. **Use the Series Expansion**:
The Taylor series for \( e^{-x} \) is given by:
\[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots \]
By substituting \( x = \log_e 2 \), we have:
\[ e^{-\log_e 2} = 1 - \log_e 2 + \frac{(\log_e 2)^2}{2!} - \frac{(\log_e 2)^3}{3!} + \ldots \]

3. **Simplify the Exponential**:
We know from properties of logarithms that:
\[ e^{-\log_e 2} = \frac{1}{e^{\log_e 2}} = \frac{1}{2} \]

4. **Conclusion**:
Therefore, the value of the original expression is:
\[ \frac{1}{2} \]

### Final Answer:
The value of \( 1 - \log_e 2 + \frac{(\log_e 2)^2}{2!} - \frac{(\log_e 2)^3}{3!} + \ldots \) is \( \frac{1}{2} \).