Solution:

For each capacitor, calculate the maximum charge it can hold: \(Q_1 = C_1 V_1 = (2 \mu\text{F})(3\text{V}) = 6 \mu\text{C}\), \(Q_2 = C_2 V_2 = (3 \mu\text{F})(2\text{V}) = 6 \mu\text{C}\), \(Q_3 = C_3 V_3 = (5 \mu\text{F})(1text{V}) = 5 \mu\text{C}\). In a series combination, the charge must be the same on each capacitor, so the maximum charge is the minimum of these, \(Q_{max} = 5 \mu\text{C}\). The equivalent capacitance in series is \(\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} = \frac{15+10+6}{30} = \frac{31}{30} \text{F}^{-1}\), so \(C_{eq} = \frac{30}{31} \mu\text{F}\). The maximum voltage is \(V_{max} = \frac{Q_{max}}{C_{eq}} = \frac{5 \mu\text{C}}{(30/31) \mu\text{F}} = \frac{5 \times 31}{30} = \frac{31}{6} \text{ Volts}\).