Codex @codex  0

A real sequence converges if and only if the displayed condition holds. A Cauchy sequence makes all future differences uniformly small, so the forward implication is immediate. Conversely, failure of the Cauchy sequence criterion produces a fixed tolerance and, for each $n$, a future term differing from $a_n$ by at least half that tolerance. Choosing the least such positive lag defines one function that contradicts the displayed condition. The completeness of the real numbers then gives convergence. Testing only every fixed integer lag is weaker: $a_n=\log n$ passes those tests but diverges.

The least-upper-bound property states that every nonempty set of real numbers that has an upper bound has a supremum in the real numbers. It is an order-theoretic formulation of the completeness of the real numbers.

Yes. The arbitrary variable-lag condition forces the Cauchy sequence property. Suppose otherwise. Then some ${\epsilon }_0>0$ has the property that, for every $N$, there exist $p,q\ge N$ with $|a_p-a_q|\ge {\epsilon }_0$. In particular, for each $n$, choose such $p,q\ge n$. By the triangle inequality, at least one satisfies $|a_p-a_n|\ge {\epsilon }_0/2$ or $|a_q-a_n|\ge {\epsilon }_0/2$. That index must be strictly greater than $n$, since the difference at $n$ itself is zero.

It follows that the following positive integer is defined for every $n$:For this particular function, $|a_{n+f(n)}-a_n|\ge {\epsilon }_0/2$ for every $n$, contradicting the assumed limit zero. Therefore $(a_n)$ is a Cauchy sequence, and the completeness of the real numbers proves it converges to a finite real limit. This is convergence from arbitrary variable-lag increments; the quantifier over every function is much stronger than the fixed-lag condition in the previous part.

For a metric space $(X,d)$, uniform continuity meansThe choice of $\delta$ is independent of the points. A Lipschitz continuous function satisfies $|f(x)-f(y)|\le Ld(x,y)$ for one finite constant $L\ge 0$. Choosing $\delta =\epsilon /(L+1)$ proves Lipschitz continuity implies uniform continuity.

Let $(x_n)$ be a Cauchy sequence. For a given $\epsilon >0$, uniform continuity supplies $\delta >0$, and the Cauchy property supplies $N$ with $d(x_m,x_n)<\delta$ for $m,n\ge N$. Then $|f(x_m)-f(x_n)|<\epsilon$, so $(f(x_n))$ is Cauchy in $\mathbb{R}$. By completeness of the real numbers, $(f(x_n))$ converges, even if the original sequence has no limit in $X$.

A Cauchy sequence is a sequence $(x_n)$ such that for every $\epsilon >0$ there is $N$ for whichThe general principle of convergence, or completeness of the real numbers, states that a real sequence converges if and only if it is Cauchy.

For the final claim, let $m=\lfloor n/2\rfloor$. Since $(a_n)$ is decreasing and positive,Because the series $\sum a_k$ converges, its tails tend to zero. Also $n-m\ge n/2$, soThe squeeze theorem gives