Using Boltzmann distribution, what is the ratio of probabilities of two states?

I got the probability of state $$i$$ (in terms of Boltzmann distribution) as $$p_{i}=\frac{1}{Z_{i}}e^{-\epsilon _{i}/{kT}},$$ where $$Z_{i}$$ is the canonical partition function: $$Z_{i}=\sum_{i}e^{-\epsilon _{i}/{kT}}$$

So the probability of state $$j$$ is $$p_{j}=\frac{1}{Z_{j}}e^{-\epsilon _{j}/{kT}},$$ where $$Z_{j}$$ is the canonical partition function: $$Z_{j}=\sum_{j}e^{-\epsilon _{j}/{kT}}$$

The ratio of the two becomes

$$\dfrac{p_{i}}{p_{j}}=\dfrac{Z_{j}}{Z_{i}}e^{-(\epsilon _{i}-\epsilon _{j})/{kT}}$$

which agrees with the correct result only if $$\frac{Z_{j}}{Z_{i}}=1$$. How to show that? Or am I missing something?

$$Z_{i}=\sum_{i}e^{-\frac{\varepsilon_{i}}{K_{\rm B}T}}$$
You sum over $$i$$, and thus $$Z$$ doesn't depend on any index. You can see this by writing explicitly
$$\sum_{i}e^{-\frac{\varepsilon_{i}}{K_{\rm B}T}}=e^{-\frac{\varepsilon_{1}}{K_{\rm B}T}}+e^{-\frac{\varepsilon_{2}}{K_{\rm B}T}}+e^{-\frac{\varepsilon_{3}}{K_{\rm B}T}}+\dots$$
There isn't any $$i$$ on the right-hand-side. This is thus simply $$Z$$ and the ratio you have is $$1$$.
The canonical partition function has no indices since the sum if over all possible states of the system. That means that in your example $$Z_i=Z_j\equiv Z$$, in which case the ratio of probabilities of two different states $$i$$ and $$j$$ is: $$\frac{p_i}{p_j}=e^{-(\epsilon_i-\epsilon_j)/kT}$$