`1-log2+((log2)^2)/(2!) -(log2)^3/(3!)+....oo`

Options

A. 2
B. `1/2`
C. `log 3`
D. `log 4`

Answer

To solve the series \( S = 1 - \log 2 + \frac{(\log 2)^2}{2!} - \frac{(\log 2)^3}{3!} + \ldots \), we can recognize that it resembles the Taylor series expansion for the exponential function. The Taylor series for \( e^x \) is given by:

\[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

However, in our case, we have a series with alternating signs. This suggests that we can use the series for \( e^{-x} \):

\[ e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots \]

### Step 1: Identify \( x \)
In our series, we can set \( x = \log 2 \). Thus, we rewrite the series:

\[ S = e^{-\log 2} \]

### Step 2: Simplify \( e^{-\log 2} \)
Using the property of logarithms and exponentials, we know that:

\[ e^{-\log 2} = \frac{1}{e^{\log 2}} = \frac{1}{2} \]

### Step 3: Conclusion
Thus, the value of the series \( S \) is:

\[ S = \frac{1}{2} \]

### Final Answer
Therefore, the value of the series \( 1 - \log 2 + \frac{(\log 2)^2}{2!} - \frac{(\log 2)^3}{3!} + \ldots \) is \( \frac{1}{2} \).

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