with $(a_n)$ and $(b_n)$ two non-negative sequences where $(b_n)$ converges to $0$ and $0 \leq a_{n+1} \leq b_n + a_n$, does $a_n$ converge?

Question

Let $(a_n)$ and $(b_n)$ be two sequences of non-negative real numbers such that for all $n \in \mathbb{N}$, $a_{n+1} \leq a_n + b_n$ and $\sum_{n=1}^{\infty} b_n$ converges. Then, $\lim_{n \to \infty} a_n$ exists and is finite.

the only thing i could think of was the same as someone did in this question, How to prove the sequence $\{b_n\}$ with $0 \leq b_{n+1} \leq b_n + a_n$ converges, where $a_n \geq 0$ converges to $0$. which is just bounding the sequence. I guess he wasn't looking for a proof of the statement so here I am.

Answer 1

Let $(a_n),(b_n)$ be nonnegative sequences such that $a_{n+1}\leq a_n+b_n$, we have

1. If $b_n\to0$, then $a_n$ may not converge (your link already provided a counterexample)
2. If $\sum_nb_n$ converges, then $a_n$ converges.

To prove 2, assume $a_n$ diverges, then there exist $\varepsilon>0$ and a sequence $(n_1,n_2,...)$ such that $|a_{n_{k+1}}-a_{n_k}|>\varepsilon$, or equivalently, $$(1)\ 0\leq a_{n_{k+1}}<a_{n_k}-\varepsilon\quad\text{or}\quad(2)\ a_{n_{k+1}}>a_{n_k}+\varepsilon.$$ Note that $(1)$ can happen finitely many times, because otherwise $a_{n_k}$ will become negative eventually.

As for $(2)$, since $\sum_nb_n$ converges it is Cauchy and therefore there exists $N\in\mathbb Z^+$ such that $\sum_{n=p}^qb_n<\varepsilon$ for all $q>p>N$. This implies $$a_{q+1}=a_p+\sum_{n=p}^q(a_{n+1}-a_n)\leq a_p+\sum_{n=p}^qb_n\leq a_p+\varepsilon.$$ We can choose $(p,q+1)$ as $(n_k,n_{k+1})$ for a sufficiently large $k$, which means $(2)$ must also happen finitely many times. But this means $(n_k)$ is finite, contradiction.