Lower current but higher potential?

The parallel combination of R2 and R3 is in series with R1. When R3 increases, the resistance of the parallel combination of R2 and R3 increases. By voltage division between the parallel combination and R1, the voltage across R2 and R3 increases. The current through R2 correspondingly increases. The current through the battery decreases.

Method 1:

Current $I$ through $R_{2}$ is represented by $\frac {R_{3}}{R_{2}R_{3}+R_{1}(R_{2}+R_{3})}$ and $\frac {dI}{dR_{3}}>0$ for all values of $R_{3}$

Method 2 (about limits, not very convincing):

Say $R_{3}$ approaches 0, the voltage across $R_{2}$ approaches 0, same for the current in $R_{2}$ too. When $R_{3}$ approaches $\infty$ , the circuit is equivalent to removing $R_{3}$ and the voltage across $R_{2}$ is $\frac {R_{2}}{R_{2}+R_{3}}$ . Hence, the current increases when $R_{3}$ increases. This method assumes that $\frac {dI}{dR_{3}}>0$ is always positive from 0 to infinity and that there are no bumps in the curve within that range.