`(1)/(2!) + (1+2)/(3!) + (1+2+3)/(4!) + (1+2+3+4)/(5!) +.... = `

Options

A. `e//3`
B. `e//4`
C. `e//2`
D. none

Answer

To solve the given series

\[ S = \frac{1}{2!} + \frac{1+2}{3!} + \frac{1+2+3}{4!} + \frac{1+2+3+4}{5!} + \ldots \]

we will first identify the general term of the series.

### Step 1: Identify the nth term
The nth term can be expressed as:

\[ T_n = \frac{1 + 2 + 3 + \ldots + n}{(n+1)!} \]

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 \(T_n\):

\[ T_n = \frac{\frac{n(n+1)}{2}}{(n+1)!} \]

### Step 2: Simplify the nth term
Now, simplifying \(T_n\):

\[ T_n = \frac{n(n+1)}{2(n+1)!} = \frac{n}{2 \cdot n!} \]

### Step 3: Write the series in summation form
Now, we can express the entire series \(S\) as:

\[ S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{n}{2 \cdot n!} \]

### Step 4: Factor out the constant
Factoring out the constant \(\frac{1}{2}\):

\[ S = \frac{1}{2} \sum_{n=1}^{\infty} \frac{n}{n!} \]

### Step 5: Simplify the summation
We know that:

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

Thus, we can rewrite the summation:

\[ S = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(n-1)!} \]

### Step 6: Change the index of summation
Let \(m = n - 1\), then as \(n\) goes from \(1\) to \(\infty\), \(m\) goes from \(0\) to \(\infty\):

\[ S = \frac{1}{2} \sum_{m=0}^{\infty} \frac{1}{m!} \]

### Step 7: Recognize the series
The series \(\sum_{m=0}^{\infty} \frac{1}{m!}\) is the Taylor series expansion for \(e^x\) at \(x=1\):

\[ \sum_{m=0}^{\infty} \frac{1}{m!} = e \]

### Step 8: Final result
Thus, substituting back, we find:

\[ S = \frac{1}{2} e \]

The final answer is:

\[ S = \frac{e}{2} \]