# Sequence $(a_n)$ converges iff sequence containing supremum of width terms converges

Let $(a_n)_{n\geq 1}$ be a complex sequence. Define the sequence $(d_n)_{n\geq 1}$ by $$d_n=\sup_{x,y\in A_n}|x-y|$$ where $A_n=\{a_k:k>n,k\in \mathbb{N}\}$. The question is to prove that $(a_n)$ converges if and only if $(d_n)$ converges to $0$.

If I look at the sequence $(a_n)$ defined by $a_n=1/n$ for $n\geq 1$, then $(d_n)$ surely is a tail of $(a_n)$. Probably it is in general.

Suppose that $(a_n)$ converges. Then it is a Cauchy sequence. Given $\epsilon>0$, there is $N\in\mathbb{N}$ such that $|a_n-a_m|<\epsilon$ for each $n,m\geq N$. However, $|d_n-0|\leq |a_n-a_m|$ for all $n,m\geq N$, this implies that $(d_n)$ converges to $0$. This approach seems like I am missing something because of the last inequality. I do not know how to prove it conversely.

• The same argument shows that $(a_n)$ is a Cauchy sequence if and only if $d_n \to 0$. – Rigel May 20 '17 at 12:43
• @Rigel This looks useful. I tried to prove it, but the notation in defining $d_n$, especially the subindex $n$, confuses the approach for proving. Could you please show any further hints how to prove it? – Hopeless May 21 '17 at 0:01

Observe that, by definition, $$d_n := \sup\{|a_k - a_j|: k, j > n\}.$$ Furthermore, it is easily seen that $d_n$ is finite if and only if $(a_n)$ is a bounded sequence and, in such case, the sequence $(d_n)$ is non-increasing.
Let us prove that $(a_n)$ is a Cauchy sequence if and only if $d_n \to 0$.
Indeed, given $\epsilon > 0$, the condition $$|a_k - a_j| \leq \epsilon \qquad \forall k,j> N$$ is equivalent to $$d_N \leq \epsilon.$$ Since $(d_n)$ is non-increasing, the last condition is equivalent to $$d_n \leq \epsilon \qquad \forall n\geq N.$$
• Simply, if $n < m$ the set of indices $k,j > n$ contains the set $k,j > m$, hence $d_n \geq d_m$. – Rigel May 21 '17 at 16:08