`(1)/(2)((1)/(5)+(1)/(7))-(1)/(4)((1)/(5^(2))+(1)/(7^(2)))+(1)/(6)((1)/(5^(3))+(1)/(7^(3)))-….oo=`

Options

A. `(1)/(2)log_(e )(5//7)`
B. `(1)/(2)log_(e )(48//35)`
C. `log_(e )(47//35)`
D. `log_(e )(35//47)`

Answer

To solve the given series

\[ S = \frac{1}{2} \left( \frac{1}{5} + \frac{1}{7} \right) - \frac{1}{4} \left( \frac{1}{5^2} + \frac{1}{7^2} \right) + \frac{1}{6} \left( \frac{1}{5^3} + \frac{1}{7^3} \right) - \ldots \]

we can break it down into two separate series, one involving \( \frac{1}{5^n} \) and the other involving \( \frac{1}{7^n} \).

### Step 1: Rewrite the series

We can express the series as:

\[ S = \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \left( \frac{1}{5^n} + \frac{1}{7^n} \right) \]

### Step 2: Separate the series

This can be separated into two parts:

\[ S = \frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \cdot 5^n} + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \cdot 7^n} \right) \]

### Step 3: Recognize the series as logarithmic

The series \( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n x^n} \) converges to \( -\ln(1 + \frac{1}{x}) \) for \( |x| > 1 \).

Thus, we can rewrite our sums:

\[ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \cdot 5^n} = -\ln\left(1 + \frac{1}{5}\right) = -\ln\left(\frac{6}{5}\right) \]

\[ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \cdot 7^n} = -\ln\left(1 + \frac{1}{7}\right) = -\ln\left(\frac{8}{7}\right) \]

### Step 4: Combine the logarithmic results

Now substituting back into our expression for \( S \):

\[ S = \frac{1}{2} \left( -\ln\left(\frac{6}{5}\right) - \ln\left(\frac{8}{7}\right) \right) \]

Using the property of logarithms \( \ln(a) + \ln(b) = \ln(ab) \):

\[ S = \frac{1}{2} \left( -\ln\left(\frac{6 \cdot 8}{5 \cdot 7}\right) \right) = -\frac{1}{2} \ln\left(\frac{48}{35}\right) \]

### Step 5: Final simplification

This can be rewritten as:

\[ S = \frac{1}{2} \ln\left(\frac{35}{48}\right) \]

Thus, the final answer is:

\[ S = \frac{1}{2} \ln\left(\frac{48}{35}\right) \]