Yes, you are correct! The correct answer to the series

$\sum_{n=0}^{\infty} \frac{(-1)^n}{n! \cdot 2^n}$

is indeed $\frac{1}{\sqrt{e}}$.

Here's the correct reasoning:

The series can be rewritten as:

$\sum_{n=0}^{\infty} \frac{(-1)^n}{n! \cdot 2^n} = \sum_{n=0}^{\infty} \frac{\left(-\frac{1}{2}\right)^n}{n!}$

This is the Taylor series expansion of the exponential function evaluated at $-\frac{1}{2}$:

$e^{-\frac{1}{2}} = \sum_{n=0}^{\infty} \frac{\left(-\frac{1}{2}\right)^n}{n!}$

However, because of the $(-1)^n$ factor, this can also be recognized as the series for $e^{-x}$ evaluated at $x = \frac{1}{2}$:

$\sum_{n=0}^{\infty} \frac{\left(-\frac{1}{2}\right)^n}{n!} = e^{-\frac{1}{2}}$

Thus, the series sums to $e^{-\frac{1}{2}}$, which is equivalent to:

$\frac{1}{\sqrt{e}}$

So the final correct value of the series is indeed $\frac{1}{\sqrt{e}}$.

Would you like to explore more details on how series expansions work or any other questions?

Here are some related questions:

1. How does the square root function impact the convergence of the series?
2. Can you generalize this result for other similar series?
3. What are some real-world applications of exponential functions like $e^{-x}$?
4. How can this series be derived directly from known series expansions?
5. What are the implications of the $\frac{1}{\sqrt{e}}$ result in probability theory?

Tip: Always check the form of a series against known expansions, like those of the exponential function, as it often simplifies the problem significantly.