The sum of the series `(1)/(2.3) + (1)/(4.5) + (1)/(6.7) + ...oo=`

Options

A. `log (2e)`
B. `log (e//2)`
C. `log (4//e)`
D. none

Answer

To solve the series \( S = \frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 5} + \frac{1}{6 \cdot 7} + \ldots \) up to infinity, we can follow these steps:

### Step 1: Rewrite the terms in the series
We can express each term in the series in a more manageable form. The general term can be written as:
\[ \frac{1}{(2n)(2n+1)} \quad \text{for } n = 1, 2, 3, \ldots \]
Thus, the series can be rewritten as:
\[ S = \sum_{n=1}^{\infty} \frac{1}{(2n)(2n+1)} \]

### Step 2: Simplify the term
We can simplify the term:
\[ \frac{1}{(2n)(2n+1)} = \frac{1}{2n} - \frac{1}{2n+1} \]
This gives us:
\[ S = \sum_{n=1}^{\infty} \left( \frac{1}{2n} - \frac{1}{2n+1} \right) \]

### Step 3: Write the series in expanded form
Now, we can expand the series:
\[ S = \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \left( \frac{1}{6} - \frac{1}{7} \right) + \ldots \]
This series is a telescoping series.

### Step 4: Identify the pattern
In this series, we can see that most terms will cancel out:
\[ S = \frac{1}{2} + \left( -\frac{1}{3} + \frac{1}{4} \right) + \left( -\frac{1}{5} + \frac{1}{6} \right) + \ldots \]
As \( n \) approaches infinity, the negative terms will cancel with the positive terms from the next fractions.

### Step 5: Evaluate the limit
The remaining terms will converge to:
\[ S = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots \]
This is equivalent to:
\[ S = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} - \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{2n+1} \]
The first series diverges, but we can analyze the logarithmic behavior.

### Step 6: Use logarithmic properties
We can relate this series to the logarithmic series:
\[ S = \log(2) \]

### Final Result
Thus, the sum of the series is:
\[ S = \log(2) \]