If `S = (1)/(1.2) - (1)/(2.3) + (1)/(3.4) -(1)/(4.5) +...oo`

Options

A. `2 log 2-1`
B. `3 log 2 `
C. `log(4//e)`
D. None

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 express each term in a more manageable form.

### Step 1: Rewrite the terms
We can rewrite the general term of the series as follows:

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

Thus, the series can be rewritten 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 the series is alternating and telescoping. Most terms will cancel out:

\[ S = 1 - \left( \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + \frac{1}{4} - \frac{1}{4} + \ldots \right) \]

Thus, we can see that:

\[ S = 1 - \lim_{n \to \infty} \frac{1}{n+1} \]

### Step 4: Evaluate the limit
As \( n \) approaches infinity, \( \frac{1}{n+1} \) approaches 0. Therefore, we have:

\[ S = 1 - 0 = 1 \]

### Final Result
Thus, the sum of the series is:

\[ S = 1 \]