The sum of the infinite series
`(1)/(2) ((1)/(3) + (1)/(4)) - (1)/(4)((1)/(3^(2)) + (1)/(4^(2))) + (1)/(6) ((1)/(3^(3)) + (1)/(4^(3)))`- ...is equal to

Options

A. `(1)/(2) log 2`
B. `log (3)/(5)`
C. `log(5)/(3)`
D. `(1)/(2) log(5/3)`

Answer

To find the sum of the infinite series

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

we can analyze the series term by term.

### Step 1: Identify the pattern in the series

The series can be rewritten as:

\[ S = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)} \left( \frac{1}{3^n} + \frac{1}{4^n} \right) \]

### Step 2: Separate the series

We can separate the series into two parts:

\[ S = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)} \frac{1}{3^n} + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)} \frac{1}{4^n} \]

Let’s denote these two sums as \( S_1 \) and \( S_2 \):

\[ S_1 = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)} \frac{1}{3^n} \]
\[ S_2 = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)} \frac{1}{4^n} \]

### Step 3: Evaluate \( S_1 \) and \( S_2 \)

Using the formula for the sum of an alternating series, we can evaluate \( S_1 \) and \( S_2 \):

The general formula for the sum of the series is:

\[ \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = -\ln(1-x) + \frac{x}{1-x} \]

For \( S_1 \) with \( x = \frac{1}{3} \):

\[ S_1 = -\ln\left(1 - \frac{1}{3}\right) + \frac{\frac{1}{3}}{1 - \frac{1}{3}} = -\ln\left(\frac{2}{3}\right) + \frac{1/3}{2/3} = -\ln\left(\frac{2}{3}\right) + \frac{1}{2} \]

For \( S_2 \) with \( x = \frac{1}{4} \):

\[ S_2 = -\ln\left(1 - \frac{1}{4}\right) + \frac{\frac{1}{4}}{1 - \frac{1}{4}} = -\ln\left(\frac{3}{4}\right) + \frac{1/4}{3/4} = -\ln\left(\frac{3}{4}\right) + \frac{1}{3} \]

### Step 4: Combine \( S_1 \) and \( S_2 \)

Now, we combine \( S_1 \) and \( S_2 \):

\[ S = S_1 + S_2 = \left(-\ln\left(\frac{2}{3}\right) + \frac{1}{2}\right) + \left(-\ln\left(\frac{3}{4}\right) + \frac{1}{3}\right) \]

### Step 5: Simplify

Combine the logarithmic terms:

\[ S = -\ln\left(\frac{2}{3} \cdot \frac{3}{4}\right) + \left(\frac{1}{2} + \frac{1}{3}\right) \]

Calculating \( \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \).

Now simplify the logarithmic term:

\[ \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2} \Rightarrow -\ln\left(\frac{1}{2}\right) = \ln(2) \]

Thus, we have:

\[ S = \ln(2) + \frac{5}{6} \]

### Final Answer

The sum of the infinite series is:

\[ S = \frac{5}{6} + \ln(2) \]