The given series is:

$\frac{1}{4e^2} - \frac{1}{8e^3} + \frac{1}{16e^4} - \frac{1}{32e^5} + \dots$

We can identify that this is an infinite geometric series where the terms alternate in sign.

Step 1: General term of the series

The series can be written as: $S = a - ar + ar^2 - ar^3 + \dots$

Here:

• $a = \frac{1}{4e^2}$ (first term),
• $r = -\frac{1}{2e}$ (common ratio).

Step 2: Sum of an infinite geometric series

The sum $S$ of an infinite geometric series is given by:

$S = \frac{a}{1 - r} \quad \text{where } |r| < 1.$

Step 3: Substitute values

We substitute $a = \frac{1}{4e^2}$ and $r = -\frac{1}{2e}$ into the formula:

$S = \frac{\frac{1}{4e^2}}{1 - \left(-\frac{1}{2e}\right)}.$

Simplify the denominator:

$1 - \left(-\frac{1}{2e}\right) = 1 + \frac{1}{2e}.$

Thus:

$S = \frac{\frac{1}{4e^2}}{1 + \frac{1}{2e}}.$

Step 4: Simplify further

To simplify $1 + \frac{1}{2e}$, we find a common denominator:

$1 + \frac{1}{2e} = \frac{2e + 1}{2e}.$

Now rewrite $S$:

$S = \frac{\frac{1}{4e^2}}{\frac{2e + 1}{2e}}.$

Dividing by a fraction is equivalent to multiplying by its reciprocal:

$S = \frac{1}{4e^2} \cdot \frac{2e}{2e + 1}.$

Simplify:

$S = \frac{2e}{4e^2 (2e + 1)}.$

Simplify the numerator and denominator:

$S = \frac{1}{2e (2e + 1)}.$

Final Answer:

The sum of the series is:

$\boxed{\frac{1}{2e (2e + 1)}}.$

Let me know if you would like me to explain any step in more detail!

Related Questions:

1. What is the formula for the sum of an infinite geometric series?
2. How do you find the common ratio of a series?
3. What is the condition for convergence of an infinite geometric series?
4. Can you solve a similar series where the ratio is not alternating?
5. What happens if the common ratio $r$ is greater than or equal to $1$?

Tip: Always check whether the common ratio $|r| < 1$ before applying the infinite geometric series formula.