Let $(a_n)$ be a sequence of non-negative real numbers, not necessarily increasing. The Cauchy criterion states that $(a_n)$ converges if and only if $\forall \epsilon >0$, there exists an $N$ such that $m >n > N$ implies that $|a_m – a_n | < \epsilon$.
Suppose one can show that for all $\epsilon > 0$, there is an $N$ such that $m > n > N$ implies $a_m – a_n < \epsilon$ (i.e. without the absolute value). Is this knowledge sufficient to conclude that $(a_n)$ converges? If not, what is a counter-example?
Best Answer
It is true. We prove it by contradiction. Suppose that $(a_n)_{n=1}^\infty$ is a sequence of non-negative real numbers such that for any $\epsilon>0$, there exists $N$ such that for all $m>n>N$, $a_m-a_n<\epsilon$. Seeking a contradiction, assume the sequence does not converge. Then there exists a number $\delta>0$ and a subsequence $(b_n)_{n=1}^\infty$ of $(a_n)_{n=1}^\infty$ such that $|b_m-b_n|>\delta$ for all $m>n$. By passing to yet another subsequence, we may assume that $(b_n)_{n=1}^\infty$ is either increasing or decreasing. If it is decreasing, then $b_{n+1}+\delta<b_n$ for all $n$. This implies that $$b_1>b_2+\delta >b_3+2\delta >\ldots >b_{n+1}+n\delta.$$ For large enough $n$, this would imply that $b_{n+1}$ is negative. In particular, it is true for any $n$ so large that $n\delta>b_1$. Thus the sequence cannot be decreasing.
But the sequence also cannot be increasing. If it were, that would mean that $b_p>b_q+\delta$ for all $p>q$. But by the original hypotheses, there exists $N$ such that for all $m>n>N$, $a_m-a_n<\delta$. Then we can fix $p>q>N$ and let $m>n>N$ be such that $b_p=a_m$ and $b_q=a_n$ (we can do this, since $(b_i)_{i=1}^\infty$ is a subsequence of $(a_i)_{i=1}^\infty$). Then $b_p-b_q>\delta$ by the properties of $(b_i)_{i=1}^\infty$, but $b_p-b_q=a_m-a_n<\delta$ by the properties of $(a_i)_{i=1}^\infty$ and $N$.