The series in question is:

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

This is a convergent series and can be simplified by recognizing that it resembles the Taylor series expansion for the exponential function $e^x$, which is:

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

If we set $x = -\frac{1}{2}$, the series becomes:

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

The given series has an additional factor of $(-1)^n$ which is a sign alternation:

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

So, the series evaluates to $e^{-1}$, which is approximately $\frac{1}{e}$.

Would you like more details on how this series is related to the exponential function or any other questions?

Here are some related questions you might consider:

1. How can the Taylor series expansion be used to approximate other functions?
2. What are the convergence properties of the series for different values of $x$ in $e^x$?
3. What is the significance of the factor $2^n$ in the original series?
4. How can the value of $e^{-1}$ be estimated numerically?
5. What other series expansions are related to the exponential function?

Tip: Understanding the relationship between series and functions like $e^x$ can greatly simplify the evaluation of complex series.