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

Options

A. `log_(e )2`
B. `log_(e )3`
C. `log_(e )5`
D. `log_(e )2-(1)/(2)`

Answer

To solve the series \( S = \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{3 \cdot 4 \cdot 5} + \frac{1}{5 \cdot 6 \cdot 7} + \ldots \), we will first identify a general term for the series and then simplify it.

### Step 1: Identify the general term
The general term of the series can be expressed as:
\[ S_n = \frac{1}{(2n-1)(2n)(2n+1)} \]
where \( n \) starts from \( 1 \).

### Step 2: Simplify the general term
We can simplify \( S_n \) using partial fractions:
\[ S_n = \frac{1}{(2n-1)(2n)(2n+1)} = \frac{A}{2n-1} + \frac{B}{2n} + \frac{C}{2n+1} \]
Multiplying through by the denominator \((2n-1)(2n)(2n+1)\) gives:
\[ 1 = A(2n)(2n+1) + B(2n-1)(2n+1) + C(2n-1)(2n) \]
Expanding this will allow us to find \( A \), \( B \), and \( C \).

### Step 3: Find coefficients
To find \( A \), \( B \), and \( C \), we can substitute convenient values for \( n \):
1. Let \( n = \frac{1}{2} \) to eliminate \( B \) and \( C \).
2. Let \( n = 0 \) to eliminate \( A \) and \( C \).
3. Let \( n = -\frac{1}{2} \) to eliminate \( A \) and \( B \).

After solving these equations, we find:
\[ A = \frac{1}{2}, \quad B = -1, \quad C = \frac{1}{2} \]
Thus,
\[ S_n = \frac{1/2}{2n-1} - \frac{1}{2n} + \frac{1/2}{2n+1} \]

### Step 4: Rewrite the series
Now we can rewrite the series:
\[ S = \sum_{n=1}^{\infty} \left( \frac{1/2}{2n-1} - \frac{1}{2n} + \frac{1/2}{2n+1} \right) \]

### Step 5: Combine the series
This series can be combined:
\[ S = \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} + \frac{1}{2n+1} \right) - \sum_{n=1}^{\infty} \frac{1}{2n} \]
The first part is related to the harmonic series, while the second part is half of the harmonic series.

### Step 6: Evaluate the series
Using known results for the harmonic series, we can evaluate:
\[ S = \frac{1}{2} \left( \log(2) + \log(2) \right) - \frac{1}{2} \log(2) \]
This simplifies to:
\[ S = \log(2) - \frac{1}{2} \]

### Final Answer
Thus, the sum of the series is:
\[ S = \log(2) - \frac{1}{2} \]