What number does this series evaluate to: (-1)^n/(n!2^n)

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.

Learn how to evaluate the series (-1)^n / (n! * 2^n) with a step-by-step solution. Understand its relation to the exponential function and the Taylor series expansion. Suitable for advanced high school and college-level students.

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