`1 + (2)/(1.2.3) + (2)/(3.4.5) + (2)/(5.6.7) + ... = `

Options

A. `log_(3) 4`
B. `2 log_(e) 2`
C. `2 log 3`
D. None

Answer

To solve the given series \( S = 1 + \frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{3 \cdot 4 \cdot 5} + \frac{2}{5 \cdot 6 \cdot 7} + \ldots \), we will analyze the pattern and derive a closed form.

### Step 1: Identify the Pattern
The series can be rewritten in terms of factorials:
\[ S = 1 + \sum_{n=1}^{\infty} \frac{2}{(2n-1)(2n)(2n+1)} \]
This is because the denominators can be expressed as products of three consecutive integers.

### Step 2: Simplify the General Term
We can simplify the general term:
\[ \frac{2}{(2n-1)(2n)(2n+1)} = \frac{2}{2n(2n-1)(2n+1)} = \frac{2}{2n} \cdot \frac{1}{(2n-1)(2n+1)} = \frac{1}{n} \cdot \frac{1}{(2n-1)(2n+1)} \]

### Step 3: Use Partial Fraction Decomposition
Next, we can use partial fraction decomposition on the term \( \frac{1}{(2n-1)(2n+1)} \):
\[ \frac{1}{(2n-1)(2n+1)} = \frac{A}{2n-1} + \frac{B}{2n+1} \]
Multiplying through by the denominator gives:
\[ 1 = A(2n+1) + B(2n-1) \]
Setting \( n = \frac{1}{2} \) gives \( A = \frac{1}{2} \) and setting \( n = -\frac{1}{2} \) gives \( B = -\frac{1}{2} \). Thus:
\[ \frac{1}{(2n-1)(2n+1)} = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \]

### Step 4: Rewrite the Series
Substituting back into the series:
\[ S = 1 + \sum_{n=1}^{\infty} \frac{1}{n} \cdot \frac{1}{(2n-1)(2n+1)} = 1 + \sum_{n=1}^{\infty} \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \]

### Step 5: Recognize the Telescoping Series
This sum is telescoping:
\[ S = 1 + \frac{1}{2} \left( \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \right) \]
The terms cancel out, leading to:
\[ S = 1 + \frac{1}{2} \cdot 1 = 1 + \frac{1}{2} = \frac{3}{2} \]

### Step 6: Final Result
Thus, the value of the series is:
\[ S = 2 \]