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.

Explore the solution to the series \(\sum_{n=0}^{\infty} \frac{(-1)^n}{n! \cdot 2^n}\) and learn why the correct answer is \(\frac{1}{\sqrt{e}}\). This detailed analysis covers Taylor series expansions, exponential functions, and series convergence, suitable for advanced undergraduate students and enthusiasts of mathematical series.

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