The sum of the series `1/1.2-1/2.3+1/3.4-1/4.5+` ... is

Options

A. `2 log_(e) 2`
B. `log_(e) 2-1`
C. `log_(e) 2`
D. `log_(e) ((4)/(e))`

Answer

To find the sum of the series \( S = \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5} + \ldots \), we can follow these steps:

### Step 1: Rewrite the terms of the series
Each term of the series can be expressed in a form that reveals a pattern. We can rewrite the general term as:
\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]
Thus, we can express the series as:
\[ S = \left( \frac{1}{1} - \frac{1}{2} \right) - \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) - \left( \frac{1}{4} - \frac{1}{5} \right) + \ldots \]

### Step 2: Group the terms
Now, we can group the terms:
\[ S = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( -\frac{1}{2} + \frac{1}{3} \right) + \left( -\frac{1}{3} + \frac{1}{4} \right) + \left( -\frac{1}{4} + \frac{1}{5} \right) + \ldots \]

### Step 3: Simplify the series
Notice that this series is telescoping. Most terms will cancel out:
\[ S = 1 - \left( \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + \ldots \right) \]
This simplifies to:
\[ S = 1 - \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots \right) \]

### Step 4: Recognize the series as a logarithmic series
The remaining series can be recognized as a series that converges to:
\[ S = 1 - \left( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \right) = 1 - \log(2) \]
This is because the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \) converges to \( \log(2) \).

### Step 5: Final expression
Thus, we can express the sum of the series as:
\[ S = 2 \log(2) - 1 \]

### Conclusion
The sum of the series \( S = \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5} + \ldots \) is:
\[ S = 2 \log(2) - 1 \]