Math Problem Solver

\[ \frac{1}{2} + \frac{1}{8} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{41}{24} = 1.708\overline{3} \]

The sum can be computed by rewriting each term over a common denominator, where the least common denominator of \(\frac{1}{2}\), \(\frac{1}{8}\), \(\frac{1}{2}\), \(\frac{1}{3}\), and \(\frac{1}{4}\) is \(24\). \[ \begin{aligned} \frac{1}{2} + \frac{1}{8} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} &= \frac{1 \cdot 12}{2 \cdot 12} + \frac{1 \cdot 3}{8 \cdot 3} + \frac{1 \cdot 12}{2 \cdot 12} + \frac{1 \cdot 8}{3 \cdot 8} + \frac{1 \cdot 6}{4 \cdot 6} \\ &= \frac{12 + 3 + 12 + 8 + 6}{24} \\ &= \frac{41}{24} \end{aligned} \]