To solve this problem, we need to evaluate the infinite series: \(S = 1 + \frac{2}{2!} + \frac{3}{3!} + \frac{4}{4!} + \cdots \infty\). We will use the properties of exponential functions and series expansion to find the sum.
The function \(e^x\) is given by the series expansion: \(e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\).
By differentiating the series expansion of \(e^x\) with respect to \(x\), we get: \(\frac{d}{dx}(e^x) = e^x = \frac{1}{0!} + \frac{x}{1!} + \frac{x^2}{2!} + \cdots\).
Now, multiply by \(x\): \(x \cdot e^x = x + \frac{x^2}{1!} + \frac{x^3}{2!} + \cdots\).
Differentiating again with respect to \(x\) gives: \((x \cdot e^x)' = e^x + x \cdot e^x = 1 + \frac{2x}{1!} + \frac{3x^2}{2!} + \cdots\).
Substitute \(x = 1\): \((x \cdot e^x)' \big|_{x=1} = e + e \cdot 1 = 2e\).
This result matches exactly with our series: \(1 + \frac{2}{2!} + \frac{3}{3!} + \frac{4}{4!} + \cdots = 2e\).
Therefore, the infinite series evaluates to \(2e\), which is the correct answer.
Thus, the correct option is:
\(<2e>\)