`x gt0`, then the sum of the series `e^(-x)-e^(-2x)+e^(-3x)-...` i s

Options

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

Answer

To find the sum of the series \( e^{-x} - e^{-2x} + e^{-3x} - e^{-4x} + \ldots \) for \( x > 0 \), we can follow these steps:

### Step 1: Define the series
Let \( S \) be the sum of the series:
\[ S = e^{-x} - e^{-2x} + e^{-3x} - e^{-4x} + \ldots \]

### Step 2: Group the terms
We can rearrange the series to group the positive and negative terms:
\[ S = (e^{-x} + e^{-3x} + e^{-5x} + \ldots) - (e^{-2x} + e^{-4x} + e^{-6x} + \ldots) \]

### Step 3: Identify the geometric series
The first group \( e^{-x} + e^{-3x} + e^{-5x} + \ldots \) is a geometric series with the first term \( a = e^{-x} \) and common ratio \( r = e^{-2x} \). The second group \( e^{-2x} + e^{-4x} + e^{-6x} + \ldots \) is also a geometric series with the first term \( a = e^{-2x} \) and common ratio \( r = e^{-2x} \).

### Step 4: Calculate the sum of the geometric series
The sum of an infinite geometric series is given by the formula:
\[ \text{Sum} = \frac{a}{1 - r} \]
For the first series:
\[ \text{Sum}_1 = \frac{e^{-x}}{1 - e^{-2x}} \]
For the second series:
\[ \text{Sum}_2 = \frac{e^{-2x}}{1 - e^{-2x}} \]

### Step 5: Substitute back into the expression for \( S \)
Now substituting these sums back into the expression for \( S \):
\[ S = \frac{e^{-x}}{1 - e^{-2x}} - \frac{e^{-2x}}{1 - e^{-2x}} \]

### Step 6: Combine the fractions
Since both terms have the same denominator, we can combine them:
\[ S = \frac{e^{-x} - e^{-2x}}{1 - e^{-2x}} \]

### Step 7: Simplify the numerator
Factoring the numerator:
\[ S = \frac{e^{-x}(1 - e^{-x})}{1 - e^{-2x}} \]

### Step 8: Factor the denominator
The denominator can be factored as:
\[ 1 - e^{-2x} = (1 - e^{-x})(1 + e^{-x}) \]
Thus, we can rewrite \( S \) as:
\[ S = \frac{e^{-x}(1 - e^{-x})}{(1 - e^{-x})(1 + e^{-x})} \]

### Step 9: Cancel common factors
We can cancel \( (1 - e^{-x}) \) from the numerator and denominator (as long as \( x > 0 \)):
\[ S = \frac{e^{-x}}{1 + e^{-x}} \]

### Step 10: Final result
Thus, the sum of the series is:
\[ S = \frac{1}{e^{x} + 1} \]