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

Options

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

Answer

To solve the series \( S = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \ldots \) we can relate it to the exponential series.

### Step 1: Recognize the Exponential Series
The exponential function \( e^x \) can be expressed as a power series:
\[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]
If we set \( x = -1 \), we have:
\[ e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \ldots \]

### Step 2: Rearranging the Series
From the series for \( e^{-1} \), we can rearrange it to isolate the terms we are interested in:
\[ e^{-1} = 1 - \left( \frac{1}{1!} - \frac{1}{2!} + \frac{1}{3!} - \frac{1}{4!} + \ldots \right) \]
This means:
\[ e^{-1} = 1 - S \]

### Step 3: Solve for \( S \)
We can express \( S \) in terms of \( e^{-1} \):
\[ S = 1 - e^{-1} \]

### Step 4: Substitute \( e^{-1} \)
Since \( e^{-1} = \frac{1}{e} \), we can substitute this into our equation:
\[ S = 1 - \frac{1}{e} \]

### Step 5: Final Result
Thus, the value of the series is:
\[ S = 1 - \frac{1}{e} \]

### Summary
The series \( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \ldots \) converges to \( 1 - \frac{1}{e} \).