1. We start with the given inequality:
$$\frac{a}{b} > \frac{31}{17}$$
We need to find the smallest fraction $\frac{a}{b}$ such that $b < 17$.

2. Rewrite $\frac{a}{b}$ in the form:
$$\frac{a}{b} = 2 - \frac{c}{b}$$
where $c$ is a positive integer. The condition $\frac{31}{17} < \frac{a}{b}$ translates into:
$$\frac{31}{17} < 2 - \frac{c}{b}$$
Simplifying this inequality, we get:
$$\frac{31}{17} < 2 - \frac{c}{b} \implies \frac{31}{17} - 2 < -\frac{c}{b} \implies -\frac{3}{17} < -\frac{c}{b} \implies \frac{3}{17} > \frac{c}{b}$$

3. To minimize $\frac{a}{b}$, we need to maximize $\frac{c}{b}$. We analyze different values of $c$ to find the maximum $\frac{c}{b}$ under the constraint $b < 17$.

4. Consider $c = 3$:
$$\frac{c}{b} = \frac{3}{b}$$
For $\frac{3}{b} > \frac{3}{17}$, we need $b < 17$. However, since $b < 17$ is already given, this does not provide a contradiction. But we need to check if this value of $c$ provides the smallest $\frac{a}{b}$.

5. Consider $c = 2$:
$$\frac{c}{b} = \frac{2}{b}$$
For $\frac{2}{b} > \frac{3}{17}$, we need:
$$b < \frac{2 \cdot 17}{3} = \frac{34}{3} \approx 11.33$$
Hence, $b \geq 12$. The smallest possible value for $\frac{c}{b}$ in this case is:
$$\frac{2}{12} = \frac{1}{6}$$

6. Consider $c = 1$:
$$\frac{c}{b} = \frac{1}{b}$$
For $\frac{1}{b} > \frac{3}{17}$, we need:
$$b < \frac{17}{3} \approx 5.67$$
Hence, $b \geq 6$. The smallest possible value for $\frac{c}{b}$ in this case is:
$$\frac{1}{6}$$

7. Summarizing, the largest possible value for $\frac{c}{b}$ is $\frac{1}{6}$, and hence the smallest possible value for $\frac{a}{b}$ is:
$$\frac{a}{b} = 2 - \frac{1}{6} = \frac{12}{6} - \frac{1}{6} = \frac{11}{6}$$

The final answer is $\boxed{\frac{11}{6}}$.