The sum of the series `(1)/(1*2*3)+(1)/(3*4*5)+(1)/(5*6*7)+. . .` is

Options

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

Answer

To find the sum of 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 first identify the general term of the series. The general term can be expressed as:

\[ T_n = \frac{1}{(2n-1)(2n)(2n+1)} \]

### Step 1: Express the General Term using Partial Fractions

We will decompose \(T_n\) into partial fractions:

\[ \frac{1}{(2n-1)(2n)(2n+1)} = \frac{A}{2n-1} + \frac{B}{2n} + \frac{C}{2n+1} \]

Multiplying through by the denominator \((2n-1)(2n)(2n+1)\) gives:

\[ 1 = A(2n)(2n+1) + B(2n-1)(2n+1) + C(2n-1)(2n) \]

### Step 2: Solve for Coefficients A, B, and C

To find \(A\), \(B\), and \(C\), we can substitute convenient values for \(n\):

1. Let \(n = \frac{1}{2}\):
\[ 1 = A(1)(2) + 0 + 0 \Rightarrow A = \frac{1}{2} \]

2. Let \(n = 0\):
\[ 1 = 0 + B(-1)(1) + 0 \Rightarrow B = -1 \]

3. Let \(n = -\frac{1}{2}\):
\[ 1 = 0 + 0 + C(-1)(0) \Rightarrow C = \frac{1}{2} \]

Thus, we have:

\[ \frac{1}{(2n-1)(2n)(2n+1)} = \frac{1/2}{2n-1} - \frac{1}{2n} + \frac{1/2}{2n+1} \]

### Step 3: Rewrite the Series

Now substituting back into the series:

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

### Step 4: Simplify the Series

This can be rearranged as:

\[ S = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{2n-1} - \sum_{n=1}^{\infty} \frac{1}{2n} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{2n+1} \]

Notice that:

- The first and last sums are similar, and they will cancel out some terms.

### Step 5: Evaluate the Sums

The sum of the series can be evaluated using known results:

- The series \(\sum_{n=1}^{\infty} \frac{1}{n}\) diverges.
- The series \(\sum_{n=1}^{\infty} \frac{1}{2n}\) converges to \(\frac{1}{2} \log(2)\).

### Final Result

After evaluating and simplifying, we find:

\[ S = \frac{1}{2} \log(2) - \frac{1}{2} + \frac{1}{2} \log(2) \]

Thus, the final sum of the series is:

\[ S = \frac{1}{2} \log(2) - \frac{1}{2} \]