The sum of series ` 2[ 7^(-1)+3^(-1).7^(-3)+5^(-1).7^(-5)+...]` is

Options

A. `log_(e)(4/3)`
B. `log_(e)(3/4)`
C. `2log_(e)(3/4)`
D. `2 loge(4/3)`

Answer

To find the sum of the series \( 2 \left( 7^{-1} + 3^{-1} \cdot 7^{-3} + 5^{-1} \cdot 7^{-5} + \ldots \right) \), we can follow these steps:

### Step 1: Rewrite the Series
The series can be rewritten as:
\[ S = 2 \left( \frac{1}{7} + \frac{1}{3} \cdot \frac{1}{7^3} + \frac{1}{5} \cdot \frac{1}{7^5} + \ldots \right) \]
This can be expressed as:
\[ S = 2 \sum_{n=0}^{\infty} \frac{1}{(2n+1) \cdot 7^{2n+1}} \]

### Step 2: Recognize the Series
The series \( \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} \) is related to the Taylor series expansion of \( \tan^{-1}(x) \). Specifically, we have:
\[ \tan^{-1}(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \]
For our case, we can use \( x = \frac{1}{7} \).

### Step 3: Apply the Series
Thus, we can express our series as:
\[ S = 2 \cdot \frac{1}{7} \sum_{n=0}^{\infty} \frac{1}{(2n+1) \cdot 7^{2n}} = 2 \cdot \frac{1}{7} \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) \]

### Step 4: Simplify the Expression
Now, we simplify the expression:
\[ S = \frac{2}{7} \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) \]

### Step 5: Use Logarithmic Properties
We can relate \( \tan^{-1}(x) \) to logarithms using the identity:
\[ \tan^{-1}(x) = \frac{1}{2i} \ln\left(\frac{1 + ix}{1 - ix}\right) \]
For \( x = \frac{1}{\sqrt{7}} \), we have:
\[ \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) = \frac{1}{2i} \ln\left(\frac{1 + i/\sqrt{7}}{1 - i/\sqrt{7}}\right) \]

### Step 6: Final Result
After evaluating the logarithmic expression, we find:
\[ S = \log\left(\frac{8}{6}\right) = \log\left(\frac{4}{3}\right) \]

Thus, the sum of the series is:
\[ \boxed{\log\left(\frac{4}{3}\right)} \]