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In the range $n \le jp^{1/2} < n+1$, the number of $j$'s contributing to the sum is $|A|p^{-1/2} +o(p^{-1/2})$. So that portion of the sum satisfies $$ \sum_{j=np^{-1/2}}^{(n+1)p^{-1/2}} p(1-p)^j \chi_I(j)= (|A|p^{1/2}+o(p^{1/2})) (1-p)^{np^{-1/2}} ; $$ here, I can treat $(1-p)^j$ as a constant when doing the sum, at the expense of a multiplicative error $(1-p)^{p^{1/2}}$, which can be absorbed into the other error term, from the ergodic theorem.

So the whole sum equals $$ |A|p^{1/2}(1+o(1)) \sum_{n\ge 0} (1-p)^{np^{-1/2}} = (1+o(1))|A| \frac{p^{1/2}}{1-(1-p)^{p^{-1/2}}} \to |A| , $$ as desired.

In the range $n \le jp^{1/2} < n+1$, the number of $j$'s contributing to the sum is $|A|p^{-1/2} +o(p^{-1/2})$. So that portion of the sum satisfies $$ \sum_{j=np^{-1/2}}^{(n+1)p^{-1/2}} p(1-p)^j \chi_I(j)= (|A|p^{1/2}+o(p^{1/2})) (1-p)^{np^{-1/2}} ; $$ here, I can treat $(1-p)^j$ as a constant when doing the sum, at the expense of a multiplicative error $(1-p)^{p^{-1/2}}$, which can be absorbed into the other error term, from the ergodic theorem.

So the whole sum equals $$ |A|p^{1/2}(1+o(1)) \sum_{n\ge 0} (1-p)^{np^{-1/2}} = (1+o(1))|A| \frac{p^{1/2}}{1-(1-p)^{p^{-1/2}}} \to |A| , $$ as desired.

In the range $n \le jp^{1/2} < n+1$, the number of $j$'s contributing to the sum is $|A|p^{-1/2} +o(p^{-1/2})$. So that portion of the sum satisfies $$ \sum_{j=np^{-1/2}}^{(n+1)p^{-1/2}} p(1-p)^j \chi_I(j)= (|A|p^{1/2}+o(p^{1/2})) (1-p)^{np^{-1/2}} , $$ at least as long as $n\ll p^{-3/2}$, when the variation of the summand $(1-p)^j$ remains small and closer inspection shows that we can absorb this into the error term coming from the ergodic theorem. We don't really need to worry about $n\gtrsim p^{-3/2}$ since these $n$ make a negligible contribution.

So the whole sum equals $$ |A|p^{1/2}(1+o(1)) \sum_{n\ge 0} (1-p)^{np^{-1/2}} = (1+o(1))|A| \frac{p^{1/2}}{1-(1-p)^{p^{-1/2}}} \to |A| , $$ as desired.

In the range $n \le jp^{1/2} < n+1$, the number of $j$'s contributing to the sum is $|A|p^{-1/2} +o(p^{-1/2})$. So that portion of the sum satisfies $$ \sum_{j=np^{-1/2}}^{(n+1)p^{-1/2}} p(1-p)^j \chi_I(j)= (|A|p^{1/2}+o(p^{1/2})) (1-p)^{np^{-1/2}} ; $$ here, I can treat $(1-p)^j$ as a constant when doing the sum, at the expense of a multiplicative error $(1-p)^{p^{1/2}}$, which can be absorbed into the other error term, from the ergodic theorem.

So the whole sum equals $$ |A|p^{1/2}(1+o(1)) \sum_{n\ge 0} (1-p)^{np^{-1/2}} = (1+o(1))|A| \frac{p^{1/2}}{1-(1-p)^{p^{-1/2}}} \to |A| , $$ as desired.

In the range $n \le jp^{1/2} < n+1$, the number of $j$'s contributing to the sum is $|A|p^{-1/2} +o(p^{-1/2})$. So that portion of the sum satisfies $$ \sum_{j=np^{-1/2}}^{(n+1)p^{-1/2}} p(1-p)^j \chi_I(j)= (|A|p^{1/2}+o(p^{1/2})) (1-p)^{np^{-1/2}} . $$ So the whole sum is approximately $$ |A|p^{1/2} \sum_{n\ge 0} (1-p)^{np^{-1/2}} = |A| \frac{p^{1/2}}{1-(1-p)^{p^{-1/2}}} \to |A| , $$ as desired.

In the range $n \le jp^{1/2} < n+1$, the number of $j$'s contributing to the sum is $|A|p^{-1/2} +o(p^{-1/2})$. So that portion of the sum satisfies $$ \sum_{j=np^{-1/2}}^{(n+1)p^{-1/2}} p(1-p)^j \chi_I(j)= (|A|p^{1/2}+o(p^{1/2})) (1-p)^{np^{-1/2}} , $$ at least as long as $n\ll p^{-3/2}$, when the variation of the summand $(1-p)^j$ remains small and closer inspection shows that we can absorb this into the error term coming from the ergodic theorem. We don't really need to worry about $n\gtrsim p^{-3/2}$ since these $n$ make a negligible contribution.

So the whole sum equals $$ |A|p^{1/2}(1+o(1)) \sum_{n\ge 0} (1-p)^{np^{-1/2}} = (1+o(1))|A| \frac{p^{1/2}}{1-(1-p)^{p^{-1/2}}} \to |A| , $$ as desired.