The expression shown is:

$1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots$

This is an alternating series, which can be recognized as the Maclaurin series expansion for the exponential function $e^{-x}$ when $x = 1$.

The general Maclaurin series for $e^x$ is:

$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

For $e^{-x}$, we substitute $-x$ for $x$:

$e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots$

So, for $x = 1$, this series becomes:

$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots$

Thus, the value of the given series is $e^{-1}$, which is approximately:

$e^{-1} \approx 0.367879$

Would you like to go into further detail or have any questions?

Here are some related questions:

1. What is the Maclaurin series expansion of $\sin(x)$ and how does it differ from $e^x$?
2. How can you estimate the value of $e^{-1}$ using the first few terms of the series?
3. Can you explain how factorial terms in a series affect the convergence?
4. How does the alternating nature of this series affect its sum?
5. How do we apply the ratio test to check the convergence of an infinite series?

Tip: Understanding how series expansions represent functions can help solve problems in calculus and mathematical physics efficiently.