Solution: Note that the given sequence is an arithmetic-geometric series. Let \[ S = \sum_{n=0}^{\infty} n \cdot \left( \frac{1}{2} \right)^n. \] First of all, we will show that the given series is convergent. For that, we will use the ratio test. Consider the general term of the series: \[ a_n = n \cdot \left( \frac{1}{2} \right)^n. \] We need to find the limit \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \] We have \begin{align*} \frac{a_{n+1}}{a_n} & = \frac{(n+1) \cdot \left( \frac{1}{2} \right)^{n+1}}{n \cdot \left( \frac{1}{2} \right)^n} \\ & = \frac{n+1}{2n}. \end{align*} Taking the limit as \( n \) approaches infinity, we get \[ \lim_{n \to \infty} \frac{n+1}{2n} = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2} = \frac{1}{2}. \] Since the limit is less than $1$, the ratio test confirms that the series is convergent.

Now we assume that \[ S = 0 \cdot \left( \frac{1}{2} \right)^0 + 1 \cdot \left( \frac{1}{2} \right)^1 + 2 \cdot \left( \frac{1}{2} \right)^2 + 3 \cdot \left( \frac{1}{2} \right)^3 + \cdots \] Multiplying both sides by $\frac{1}{2}$, we get \[ \frac{S}{2} = 0 \cdot \left( \frac{1}{2} \right)^1 + 1 \cdot \left( \frac{1}{2} \right)^2 + 2 \cdot \left( \frac{1}{2} \right)^3 + 3 \cdot \left( \frac{1}{2} \right)^4 + \cdots \] Subtracting the above equation from $S$, we get \begin{align*} S - \frac{S}{2} & = \left( 1 \cdot \left( \frac{1}{2} \right)^1 + 2 \cdot \left( \frac{1}{2} \right)^2 + 3 \cdot \left( \frac{1}{2} \right)^3 + \cdots \right) \\ & \quad - \left( 0 \cdot \left( \frac{1}{2} \right)^1 + 1 \cdot \left( \frac{1}{2} \right)^2 + 2 \cdot \left( \frac{1}{2} \right)^3 + \cdots \right) \\[1ex] \implies \frac{S}{2} & = \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots \\[1ex] \implies S & = 2 \sum_{n = 1}^{\infty} \left( \frac{1}{2} \right)^n = 2 \left( \frac{\frac{1}{2}}{1-\frac{1}{2}} \right) = 2. \end{align*} Thus, \[ \textcolor{blue}{\boxed{ \sum_{n=0}^{\infty} n \cdot \left( \frac{1}{2} \right)^n = 2 }} \]