Let $p$ be a prime and $(a_n)_{n\in\mathbb N}$ be a sequence of $\mathbb Q$ that converges to $0$ for the $p$-adic valuation and such that $(v_p(a_n))_{n\in\mathbb N}$ is increasing. One assumes that $\sum_{n\ge0}a_n$ converges to $0$ in $\mathbb Q_p$.
What can say about $v_p(a_n)$?
What I propose: for all $n$, $v_p(a_n)=v_p(\sum_{l=0}^{n-1}a_l)$.
Indeed, assumes that there exists $n_0\in\mathbb N$ such that $v_p(a_{n_0})\ne v_p(\sum_{l=0}^{n_0-1}a_l)$.
Case 1: if $v_p(a_{n_0})<v_p(\sum_{l=0}^{n_0-1}a_l)$ then for all $n>n_0$ $v_p(\sum_{l=0}^na_l)=v_p(\sum_{l=0}^{n_0-1}a_l+\sum_{l=n_0}^na_l)$. Since $v_p(a_n)$ is increasing, for all $n>n_0$, $v_p(a_n)>v_(a_{n_0})$ and $v_p(\sum_{l=n_0}^na_l)=v_p(a_{n_0})$. As $v_p(a_{n_0})<v_p(\sum_{l=0}^{n_0-1}a_l)$, one deduces that $v_p(\sum_{l=0}^na_l)=v_p(a_{n_0})$ and $\sum_{l=0}^na_l$ can not converges to $0$. In a same way, one proves
Case 2: if $v_p(a_{n_0})>v_p(\sum_{l=0}^{n_0-1}a_l)$ then $v_p(\sum_{l=0}^na_l)=v_p(\sum_{l=0}^{n_0-1}a_l)$ and again $\sum_{l=0}^na_l$ can not converges to $0$.
Is my proof correct? If yes, can one say more about $v_p(a_n)$?
If $v(a_{n+1})> v(a_n)$ then $v(\sum_n a_n)=v( a_0)$ ie. $\sum_n a_n\ne 0$
If $v(a_n)$ is increasing but not strictly increasing there is not much to say since with $S=\sum_n a_n$ we can always add a term $-S$ to the sequence to obtain that the sum is $0$.