Let $(a_k)_{k=m}, (b_k)_{k=m}$ be sequences of real numbers. If $m\le p$ define $\alpha_p$ to be the infimum of the $p$th tail, and similarly for $\beta_p,\gamma_p,\delta_p,\epsilon_p,\zeta_p$.
\begin{align*} \alpha_p & = \inf(a_k)_{k=p} \\ \beta_p &= \sup(a_k)_{k=p} \\ \gamma_p &= \inf(b_k)_{k=p} \\ \delta_p &= \sup(b_k)_{k=p} \\ \epsilon_p &= \inf(a_k+b_k)_{k=p} \\ \zeta_p &= \sup(a_k+b_k)_{k=p}\end{align*}
Suppose that $A,B\in\Bbb R$ are such that
$$ \bigcap_{k=m}^\infty [\alpha_k,\beta_k] = \{A\}, \quad \bigcap_{k=m}^\infty [\gamma_k,\delta_k] = \{B\} $$
Without reference to the concept of limits, show that, for any $p\ge m$ we have $A+B\in [\epsilon_p,\zeta_p]$.
To show $\epsilon_p =\inf(a_k+b_k)_{k=p}\le A+B$ I tried assuming for contradiction that $A+B < \inf(a_k+b_k)_{k=p}$. Let $\omega \in\Bbb R^+$ such that $A+B+\omega < \inf(a_k+b_k)_{k=p}$.
It follows that, for every $p\le q$ we have
\begin{align*} A+B +\omega &< a_q+b_q \end{align*}
Presumably we want some kind of contradiction like finding $A+\omega < a_q$ for every $q\ge p$, so that therefore $A < \alpha_q$ and then $A\notin \bigcap_{k=m}^\infty [\alpha_k,\beta_k]$.
We have $A+\omega < a_q+b_q - B$ and then need $a_q+b_q - B\le a_q$ which is equivalent to $b_q\le B$. But this is not necessarily true for all $p\le q$ so this feels a bit like a dead-end.