Convergence of a series of positive numbers with the property $(\forall\delta>0)(\exists n_{0})(\forall n> n_0)(\forall k) a_{n+k}\leq a_n+\delta$

Question

Problem: Let $a_n$ be a sequence of positive real numbers with the property $$(\forall\delta>0)(\exists n_{0}(\forall n> n_0)(\forall k) a_{n+k}\leq a_n+\delta.$$ Prove that $a_n$ converges. A hint is given in the book stating that the reader should use the inequality $ a_{n+k}\leq a_n+\delta$ and apply the limit superior for k and then limit inferior for n to the new inequality. What does the hint mean, and is it possible to do this differently?

Answer 1

$\{a_n: n\in \Bbb N\}$ is a bounded set: Take $m\in \Bbb N$ such that $\forall k\in \Bbb N\,(a_{(m+1+k)}\le a_{(m+1)}+1).$ Let $M=\max (\{a_{(m+1)}+1\}\cup \{a_j:m\ge j\in \Bbb N\}).$ Then $\forall n\in \Bbb N\,(0<a_n\le M\in \Bbb R^+).$

A bounded real sequence $(a_n)_{n\in \Bbb N}$ does $not$ converge iff $\lim \inf a_n<\lim\sup a_n.$ For brevity let $A=\lim\sup a_n$ and let $B=\lim \inf a_n.$ We have $A,B \in \Bbb R$ because $0\le B\le A\le M\in \Bbb R^+.$

Suppose by contradiction that $(a_n)_{n\in \Bbb N}$ does $not$ converge. So let $A-B=3r>0.$ Then the sets $S=\{n\in \Bbb N: |a_n-A|<r\}$ and $T=\{n\in \Bbb N: |a_n-B|<r\}$ are both infinite. Also, if $n_1\in S$ and $n_2\in T$ then $a_{n_1}-a_{n_2}>r.$ But then if $\delta=r,$ it is not possible to find the $n_0$ of the Q.