Value of `1+1/(1+2)+1/(1+2+3)+....+1/(1+2+3+....+n)` is equal to

Options

A. `(2n) /(n+1)`
B. `(3n)/(2n+1)`
C. `(4n)/(3n+1)`
D. `n/(n+1)`

Answer

To find the value of the series \( S = 1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \ldots + \frac{1}{1 + 2 + 3 + \ldots + n} \), we can follow these steps:

### Step 1: Identify the general term
The general term of the series can be expressed as:
\[ T_n = \frac{1}{1 + 2 + 3 + \ldots + n} \]
The sum of the first \( n \) natural numbers is given by the formula:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]
Thus, we can rewrite \( T_n \) as:
\[ T_n = \frac{1}{\frac{n(n + 1)}{2}} = \frac{2}{n(n + 1)} \]

### Step 2: Write the series in terms of \( T_n \)
Now we can express the entire series \( S \) as:
\[ S = \sum_{n=1}^{n} T_n = \sum_{n=1}^{n} \frac{2}{n(n + 1)} \]

### Step 3: Simplify the summation
We can simplify \( \frac{2}{n(n + 1)} \) using partial fractions:
\[ \frac{2}{n(n + 1)} = \frac{2}{n} - \frac{2}{n + 1} \]
Thus, the series becomes:
\[ S = \sum_{n=1}^{n} \left( \frac{2}{n} - \frac{2}{n + 1} \right) \]

### Step 4: Expand the summation
Expanding this summation gives:
\[ S = \left( \frac{2}{1} - \frac{2}{2} \right) + \left( \frac{2}{2} - \frac{2}{3} \right) + \left( \frac{2}{3} - \frac{2}{4} \right) + \ldots + \left( \frac{2}{n} - \frac{2}{n + 1} \right) \]

### Step 5: Observe the cancellation
Notice that this is a telescoping series. Most terms will cancel out:
\[ S = 2 - \frac{2}{n + 1} \]

### Step 6: Final expression for \( S \)
Thus, we can simplify \( S \) to:
\[ S = 2 - \frac{2}{n + 1} = \frac{2(n + 1) - 2}{n + 1} = \frac{2n}{n + 1} \]

### Final Result
The value of the series is:
\[ S = \frac{2n}{n + 1} \]