The sum of the infinite series
`(1)/(3)+(3)/(3.7)+(5)/(3.7.11)+(7)/(3.7.11.15)+"…………"` is

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. `(1)/(6)`
D. `(1)/(4)`

Answer

To find the sum of the infinite series

\[ S = \frac{1}{3} + \frac{3}{3 \cdot 7} + \frac{5}{3 \cdot 7 \cdot 11} + \frac{7}{3 \cdot 7 \cdot 11 \cdot 15} + \ldots \]

we will derive the general term and then find the sum.

### Step 1: Identify the nth term of the series

The numerators of the series are odd numbers: 1, 3, 5, 7, ... which can be expressed as \(2n - 1\) for \(n = 1, 2, 3, \ldots\).

The denominators are products of terms starting from 3 and increasing by 4 each time:
- For \(n=1\), the denominator is \(3\).
- For \(n=2\), the denominator is \(3 \cdot 7\).
- For \(n=3\), the denominator is \(3 \cdot 7 \cdot 11\).
- For \(n=4\), the denominator is \(3 \cdot 7 \cdot 11 \cdot 15\).

The general term can be expressed as:

\[ T_n = \frac{2n - 1}{\prod_{k=1}^{n} (4k - 1)} \]

where \(\prod_{k=1}^{n} (4k - 1)\) represents the product of the terms in the denominator.

### Step 2: Rewrite the series

The series can be rewritten as:

\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{2n - 1}{\prod_{k=1}^{n} (4k - 1)} \]

### Step 3: Simplify the series

We can express the series in a more manageable form. Notice that:

\[ T_n = \frac{2n - 1}{(4n - 1)(4(n-1) - 1)(4(n-2) - 1) \ldots (3)} \]

### Step 4: Use telescoping series

To find the sum, we can express the series in terms of a telescoping series. We can rewrite \(T_n\) in a form that allows us to see the cancellation of terms.

\[ T_n = \frac{1}{2} \left( \frac{1}{\prod_{k=1}^{n} (4k - 1)} - \frac{1}{\prod_{k=1}^{n+1} (4k - 1)} \right) \]

### Step 5: Sum the series

Now, we can sum the series:

\[ S = \sum_{n=1}^{\infty} T_n = \frac{1}{2} \left( \frac{1}{3} + \sum_{n=2}^{\infty} \left( \frac{1}{\prod_{k=1}^{n} (4k - 1)} - \frac{1}{\prod_{k=1}^{n+1} (4k - 1)} \right) \right) \]

This is a telescoping series, which simplifies to:

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

### Step 6: Calculate the final sum

Calculating the above expression gives:

\[ S = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3} \]

Thus, the sum of the infinite series is:

\[ \boxed{\frac{1}{2}} \]