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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).

Jürgen Sukumaran
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    If $p_n$ is bounded then $e_n+a_np_{n+1}\in \ell^1$. Therefore you have $$p_{n+1}-p_n\le f_n, $$ for a $f_n\in \ell^1$. Show that this implies that the Cauchy criterion is satisfied. P.S.: Actually I see a problem here, because you should have $|p_{n+1}-p_n|\le f_n$ for the Cauchy criterion to kick in. – Giuseppe Negro Jan 15 '18 at 09:31
  • Indeed, I was trying to show it was a Cauchy sequence at first using what you describe but I would also have to show that $p_{n+1} - p_n > -\varepsilon$ – Jürgen Sukumaran Jan 15 '18 at 09:50
  • Warning: the fact that $p_{n+1} - p_n\to 0$ does NOT imply that $p_n$ converges. For that you need that the series $\sum |p_{n+1}-p_n|$ is convergent. – Giuseppe Negro Jan 15 '18 at 19:26
  • I disagree with your comment. I'll post a link to a counterexample as soon as possible. – Giuseppe Negro Jan 16 '18 at 11:41
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    https://math.stackexchange.com/q/107336/8157 – Giuseppe Negro Jan 16 '18 at 11:43
  • Ah, I see where my misunderstanding is. Thanks for the help, now it is clear why $|p_{n+1}-p_n|<\varepsilon$ is not enough. – Jürgen Sukumaran Jan 16 '18 at 12:00

1 Answers1

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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. $\square$

Giuseppe Negro
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