Let $(a_n)_{n\in\mathbb{N}}, (e_n)_{n\in\mathbb{N}}, (p_n)_{n\in\mathbb{N}}, (r_n)_{n\in\mathbb{N}}$ be nonnegative sequences in $\mathbb{R}$ with $(a_n)_{n\in\mathbb{N}}\in\ell^1$ and $(e_n)_{n\in\mathbb{N}}\in\ell^1$. Furthermore, assume that $(1-a_n)p_{n+1}-p_n+r_n\leq e_n$ holds for all $n$. It is claimed that since the above inequality can be rearranged to $p_{n+1} - p_n + r_n \leq e_n + a_np_{n+1}$ that if $p_n$ is bounded above then $p_n$ converges. Can someone explain that claim? I understand how to show that $p_n$ is bounded above but cannot see why that implies convergence?
Trying to show that the sequence $(p_n)_{n\in\mathbb{N}}$ satisfies the Cauchy critera, I did the following (assume an upper bound $M \geq 0$ for $(p_n)_{n\in\mathbb{N}}$):
$p_{n+1}-p_n \leq p_{n+1}-p_n + r_n \leq e_n + a_np_{n+1}\leq e_n+Ma_n$
with the right hand side going to $0$ as $n\rightarrow \infty$ since $a_n$ and $e_n$ are in $\ell^1$. However, I would also need to show that $p_{n+1} - p_n >-\varepsilon$ for an arbitary $\varepsilon > 0$ in order for this to imply $|p_{n+1}-p_n|< \varepsilon$ (the Cauchy criteria).
The fact that $p_n$ converges is a consequence of the following lemma.
Lemma. Suppose that $p_n\ge 0$ is such that $$\tag{1} p_{n+1}-p_n\le f_n, $$ where $f_n\ge 0$ satisfies $\sum_{n=1}^\infty f_n <\infty.$ Then $p_n$ is convergent.
Proof. It suffices to prove that $$\sum_{n=1}^\infty |p_{n+1}-p_n| <\infty.$$ We use the notation $$u^+=\max(u, 0),\quad u^-=\max(-u, 0), $$ so that $$\tag{2}|p_{n+1}-p_n|=(p_{n+1}-p_n)^+ + (p_{n+1}-p_n)^-.$$ The assumption (1) implies that $$\tag{3}\sum_{n=1}^\infty (p_{n+1}-p_n)^+ < \infty. $$ Now we observe that, setting $p_0=0$ $$ 0\le p_{n+1}=\sum_{k=0}^n (p_{k+1}-p_k) = \sum_{k=0}^n (p_{k+1}-p_k)^+ -\sum_{k=0}^n (p_{k+1}-p_k)^-, $$ so, rearranging, we have the bound $$\tag{4} \sum_{n=1}^\infty (p_{n+1}-p_n)^- \le \sum_{n=1}^\infty (p_{n+1}-p_n)^+<\infty.$$ Because of the decomposition (2), the fact that the series (3) and (4) are convergent implies that $\sum |p_{n+1}-p_n|$ is convergent, and so the proof is complete. $\square$