Let $p$ be a prime and $b \in \mathbb Q$ some rational number of $p$-adic value $\lvert b \rvert_p \le 1$. Further, let $(a_0, a_1, a_2, a_3,\dotsc)$ be a strictly increasing sequence of natural numbers. I'm studying series of the form
$$S(b, (a_n)_n) := bp^{a_1} + b^2p^{a_2}+b^3p^{a_3}+\dotsc$$
which under the above hypotheses converge in $\mathbb Q_p$.
It's easy to see that this limit is rational (i.e $\in \mathbb Q$) when the series is "periodic" in the sense that there are $j \le k \in \mathbb N$ such that $a_{i+mj} = a_i + mk$ for all $m \in \mathbb N_0$ and $0 \le i \le j-1$, i.e. $$(a_0,a_1,a_2,\dotsc ,a_0+k, a_1+k, \dotsc ,a_0+2k, a_1+ 2k ,\dotsc).$$
It's also rational if the sequence is eventually periodic: that is, periodic after some finite number of terms.
By analogy with $p$-adic expansions generally I feel that the converse must also be true -- that the limit must be irrational when the sequence $(a_n)_n$ is not eventually periodic in the above sense. Am I correct, and if so how can I prove it?
Every $p$-adic number can be written as a quotient of two $p$-adic integers, so in that sense, every $p$-adic number is “rational”. In fact, for $p > 2$, $e^p \in \Bbb{Z}_p$, even though $e^p$ is not a rational number in $\Bbb{R}$.
Edit: @TorstenSchoeneberg in the comments tried to tell me that "$e^p$" (as defined by the usual convergent series $\sum_{n \geq 0} \frac{p^n}{n!}$) is not a number that exists in both $\Bbb{R}$ and $\Bbb{Q}_p$, because the metrics are different.
But the same logic he cites renders nonsensical his (and other people's) assertion that the OP is "obviously" looking for answers in "$\Bbb{Q} \cap \Bbb{Q}_p$", since the type of infinite series OP is discussing, $$b p^{a_1} + b^2 p^{a_2} + b^3 p^{a_3} + ...$$ with $0 < a_1 < a_2 < a_3 < ...,$ does not even converge in $\Bbb{Q}$. Commenters who think the OP wants an answer in "$\Bbb{Q} \cap \Bbb{Q}_p$" (whatever that is) are wrong, and have failed to appreciate this subtle but essential distinction.