Let $a_n, b_n, c_n$ be real positive sequences. Assume $a_n\to C>0$, and that $a_n$ is not monotone on any neighborhood of $+\infty$. Suppose also the following lemmata have been proven:
Lemma 1. Whenever $a_n\ge a_{n-1}$ one has $b_n\ge c_n$.
Lemma 2. For infinitely many $n$ one has $\dfrac{a_nb_n}{c_n}< C$.
Now say we want to establish that $a_n<C$ infinitely often. Then a possible approach is showing that if $a_{n-1}-C>0$ for some $n$, then $a_n<a_{n-1}$: this implies $a_n-C$ must attain infinitely many negative values because otherwise $a_n$ would be eventually decreasing, which contradicts our hypothesis.
I would like to know if the following reasoning is sound:
given an arbitrary $n$, assume without loss of generality that $a_{n-1}>C$, and suppose for the sake of contradiction that $a_n\ge a_{n-1}$. So in particular we're assuming $a_n\ge a_{n-1}>C$. But combining Lemma 1 and Lemma 2 we get that, for some choice of $n$, we have $a_n<C\dfrac{c_n}{b_n}<C.$ This is absurd, because we could have picked such an $n$ in our initial assumptions, in which case we should have $a_n> C$. It follows that $a_{n-1}>C$ implies $a_n<a_{n-1}$.
Incorrect as you can get unlucky and the inequality in Lemma 2 holds only for $a_{n-1}$ which is lower than $a_{n} ,a_{n-2}$ so no restrictions on $b_{n-1},c_{n-1}$
$a_{2n}=C+\frac{1}{n+1}, a_{2n-1}= C+\frac{1}{2^n+100}, b_{2n}=c_{2n}=1, 100b_{2n-1}=c_{2n-1}=1$ satisfy your conditions but $a_{n}>C$