Math Problem Solver

\[ \frac{1}{2} + \frac{1}{3} + \frac{1}{7} = \frac{41}{42} = 0.9\overline{761904} \]

The sum can be computed by rewriting each term over a common denominator, where the least common denominator of \(\frac{1}{2}\), \(\frac{1}{3}\), and \(\frac{1}{7}\) is \(42\). \[ \begin{aligned} \frac{1}{2} + \frac{1}{3} + \frac{1}{7} &= \frac{1 \cdot 21}{2 \cdot 21} + \frac{1 \cdot 14}{3 \cdot 14} + \frac{1 \cdot 6}{7 \cdot 6} \\ &= \frac{21 + 14 + 6}{42} \\ &= \frac{41}{42} \end{aligned} \]