The sum of the series `log_(4)2-log_(8)2+log_(16)2- . . .` is

Options

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

Answer

To find the sum of the series \( \log_{4}2 - \log_{8}2 + \log_{16}2 - \ldots \), we can follow these steps:

### Step 1: Rewrite the logarithms in terms of base 2
We start by expressing each logarithm in the series 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, we can rewrite the series as:
\[ \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots \]

### Step 2: Identify the pattern of the series
The series can be expressed in a more general form:
\[ \sum_{n=2}^{\infty} (-1)^{n} \frac{1}{n} \]
This is an alternating series where the terms are decreasing in absolute value.

### Step 3: Recognize the series as a known series
The series \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \) converges to \( \log(2) \). However, we need to adjust our series to fit this known result.

### Step 4: Adjust the series
We can express our series starting from \( n=2 \):
\[ \sum_{n=2}^{\infty} (-1)^{n} \frac{1}{n} = -\left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots \right) \]
This can be rewritten as:
\[ -\left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} - 1 \right) = -\log(2) + 1 \]

### Step 5: Final result
Thus, the sum of the series \( \log_{4}2 - \log_{8}2 + \log_{16}2 - \ldots \) is:
\[ 1 - \log(2) \]

### Summary
The sum of the series \( \log_{4}2 - \log_{8}2 + \log_{16}2 - \ldots \) is \( 1 - \log(2) \).