`log_4 2-log_8 2+log_16 2-.....oo`

Options

A. `e^(-2)`
B. `log_(e)2+1`
C. `log_(e)3-2`
D. `1-log_(e)2`

Answer

To solve the question \( \log_4 2 - \log_8 2 + \log_{16} 2 - \ldots \) up to infinity, we will follow these steps:

### Step 1: Rewrite the logarithms in terms of base 2
Using the change of base formula, we can express the logarithms in terms of base 2:
\[ \log_4 2 = \frac{\log_2 2}{\log_2 4} = \frac{1}{2}, \quad \log_8 2 = \frac{\log_2 2}{\log_2 8} = \frac{1}{3}, \quad \log_{16} 2 = \frac{\log_2 2}{\log_2 16} = \frac{1}{4} \]
Thus, the series can be rewritten as:
\[ \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \ldots \]

### Step 2: Identify the pattern in the series
The series can be expressed as:
\[ \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n} \]
This is an alternating series.

### Step 3: Relate the series to the logarithmic expansion
The series \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = \log(2) \). Therefore, we can express our series in terms of this known series:
\[ \frac{1}{2} - \left( \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots \right) \]
The series \( \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots \) can be rewritten as:
\[ -\left(\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots \right) = -(\log(2) - 1) \]

### Step 4: Combine the results
Thus, we have:
\[ \frac{1}{2} - (-(\log(2) - 1)) = \frac{1}{2} + \log(2) - 1 \]
This simplifies to:
\[ \log(2) - \frac{1}{2} + 1 = \log(2) + \frac{1}{2} \]

### Step 5: Final expression
The final expression can be simplified to:
\[ 1 - \log(2) \]

### Conclusion
The value of the infinite series \( \log_4 2 - \log_8 2 + \log_{16} 2 - \ldots \) is:
\[ 1 - \log_2 e \]