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 using the change of base formula
Using the change of base formula, we can express the logarithms in terms of base 2:
\[ \log_{m} n = \frac{\log_{b} n}{\log_{b} m} \]
For our series, we rewrite each term:
\[ \log_{4} 2 = \frac{\log_{2} 2}{\log_{2} 4} = \frac{1}{2}, \quad \log_{8} 2 = \frac{\log_{2} 2}{\log_{2} 8} = \frac{1}{3}, \quad \log_{16} 2 = \frac{\log_{2} 2}{\log_{2} 16} = \frac{1}{4} \]

### Step 2: Substitute the values into the series
Now substituting these values into the series, we get:
\[ \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots \]

### Step 3: Identify the pattern in the series
The series can be expressed as:
\[ S = \frac{1}{2} - \left(\frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots\right) \]
This can be rewritten by factoring out the negative sign:
\[ S = \frac{1}{2} - \left(-\frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \ldots\right) \]

### Step 4: Recognize the alternating series
The series \( \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots \) can be recognized as an alternating series. We can express it in terms of the natural logarithm:
\[ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = \log(2) \]
Thus, we can write:
\[ S = \frac{1}{2} - \left(\log(2) - 1\right) \]

### Step 5: Simplify the expression
Now, simplifying the expression gives:
\[ S = 1 - \log(2) \]

### Conclusion
Thus, the sum of the series is:
\[ S = 1 - \log_{e} 2 \]