I'm posting the proof of the claim in the original question in my own words:
claim: If $ \forall n \in \mathbb{N}: a_n > 0$ and $ \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} < 1$
then
(i) $\{ a_n \}_n $ converges
(ii) $\lim a_n = 0$
proof:
$ 0 \leq \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} < 1 \implies \exists 0< s <1: \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = s$
$ \iff \forall \varepsilon > 0 \exists N: n > N \implies \frac{a_{n+1}}{a_n} \in [s-\varepsilon, s+\varepsilon]$
Choose $\varepsilon$ s.t. $s+\varepsilon < 1$ and $s-\varepsilon > 0$. Then $ 0 < s-\varepsilon \leq \frac{a_{n+1}}{a_n} \leq s+\varepsilon < 1$
$ \iff (s-\varepsilon) a_n \leq a_{n+1} \leq (s+\varepsilon) a_n$
and setting $\delta := s+\varepsilon$:
$ \implies a_{n+k} \leq \delta a_{n+k-1} \leq \delta^2 a_{n+k-2} \leq \dots (\forall n > N)$
$ \implies a_{n+k} \leq \delta^{k} a_n (\forall n > N)$
Then $0 \leq a_{n+k} \leq a_n \delta^k$ and $a_n \lim_k \delta^k = 0$ implies
$\lim_k a_{n+k} = 0$
$ \implies \lim_n a_n = 0$