Solution: Let us write a few terms of the sequence. \begin{align*} s_1 & = 0 & s_2 & = 0 \\ s_3 & = \frac{1}{2} & s_4 & = \frac{1}{2^2} \\ s_5 & = \frac{1}{2} + \frac{1}{2^2}, & s_6 & = \frac{1}{2^2} + \frac{1}{2^3} \\ s_7 & = \frac{1}{2} +\frac{1}{2^2} + \frac{1}{2^3}, & s_8 & = \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} \\ & \vdots & & \vdots \\ s_{2m+1} & = \sum_{i= 1}^m \frac{1}{2^i} & s_{2m} & = s_{2m + 1} - \frac{1}{2} \\ & = \frac{\frac{1}{2} \left( 1 - \frac{1}{2^m} \right) }{1 - \frac{1}{2}} = 1 - \frac{1}{2^m}, & s_{2m} & = \frac{1}{2} - \frac{1}{2^m}. \end{align*} Since $\displaystyle \lim_{m \to \infty} \frac{1}{2^m} = 0$, the sequence $\{ s_n \} $ has two limit points, namely, $\{ 1, \frac{1}{2} \} $. Hence, we have \[ \limsup_{n \to \infty} s_n = 1 \quad \text{ and } \quad \liminf_{n \to \infty} s_n = \frac{1}{2}. \]