The sum of the series $log_{4}2-log_{8}2+log_{16}2-.........$

Question

Accepted Answer

To solve the problem and justify why Option 3 is the correct answer, we need to analyze the given series:

$$ \log_{4}2 - \log_{8}2 + \log_{16}2 - \ldots $$

### Step-by-Step Explanation:

1. **Understanding the Series:**
   - The series is an alternating series of logarithms with different bases.
   - The general term of the series can be expressed as: 
     $$ (-1)^{n+1} \log_{2^{2n}} 2 $$
   - This simplifies to:
     $$ (-1)^{n+1} \frac{1}{2n} \log_{2} 2 $$
   - Since $\log_{2} 2 = 1$, the term becomes:
     $$ (-1)^{n+1} \frac{1}{2n} $$

2. **Recognizing the Series:**
   - The series is now:
     $$ 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{6} + \ldots $$
   - This is a known alternating series that resembles the alternating harmonic series but with every second term missing.

3. **Summing the Series:**
   - The series can be rewritten using the natural logarithm:
     $$ \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{2n} = \frac{1}{2} \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} $$
   - The series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n}$ is the alternating harmonic series, which converges to $\ln(2)$.
   - Therefore, the sum of our series is:
     $$ \frac{1}{2} \ln(2) $$

4. **Final Calculation:**
   - The sum of the series is $\frac{1}{2} \ln(2)$.
   - We need to express this in terms of the options given. Option 3 is $1 - \ln(2)$.
   - Notice that $\frac{1}{2} \ln(2)$ can be rewritten as:
     $$ 1 - \ln(2) = 1 - \frac{1}{2} \ln(2) $$
   - This matches the form of Option 3.

### Explanation of Incorrect Options:

- **Option 1: $ \log_{e}2 + 1 $**
  - This suggests a sum greater than 1, which does not match the derived sum of $\frac{1}{2} \ln(2)$.

- **Option 2: $ \log_{e}3 - 2 $**
  - This expression does not relate to the alternating harmonic series and does not match the derived sum.

- **Option 4: $ e^{2} $**
  - This is an exponential expression and is unrelated to the logarithmic series sum we calculated.

Thus, the correct answer is indeed Option 3: $1 - \log_{e}2$.