`(1)/(1*2)+(1)/(2*3)+(1)/(3*4)+"…."+(1)/(n(n+1))` equals

Options

A. `(1)/(n(n+1))`
B. `(n)/(n+1)`
C. `(2n)/(n+1)`
D. `(2)/(n(n+1))`

Answer

To solve the series

\[ S = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \ldots + \frac{1}{n(n+1)} \]

we can start by rewriting each term in the series using the method of partial fractions.

### Step 1: Rewrite the general term
The general term can be expressed as follows:

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

This is derived from the identity:

\[ \frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1} \]

By solving for \(A\) and \(B\), we find that \(A = 1\) and \(B = -1\).

### Step 2: Substitute the general term into the series
Now we can rewrite the entire series \(S\):

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

### Step 3: Observe the cancellation
Notice that in this series, many terms will cancel out:

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

### Step 4: Simplify the result
Thus, we can simplify the expression:

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

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

\[ S = \frac{n}{n+1} \]