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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?

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You can't write

$$Z_{i}=\sum_{i}e^{-\frac{\varepsilon_{i}}{K_{\rm B}T}}$$

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 $1$.

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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: $$ \frac{p_i}{p_j}=e^{-(\epsilon_i-\epsilon_j)/kT} $$

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