Why is this correct for a divergent series?

Question

I recently read something and I can't wrap my head around it:

For a given divergent sequence $a_n$, we have the following property: $$\lim \limits_{n \to\infty} |a_{n+p} - a_n| = 0$$ for a fixed value $p\in\mathbb{Q}$. To me it seems like this violates the cauchy sequence for convergent sequences, since we are considering a divergent series here. What is it that I am missing?

Answer 1

Another nice example is $a_n = \sqrt{n}$. Of course $a_n \to \infty$. But for a fixed $p$ we have as $n \to \infty$ $$ \sqrt{n+p} - \sqrt{n} = \sqrt{n}\left(\sqrt{1+\frac{p}{n}} - 1\right) = \sqrt{n}\left(1+\frac{p}{2n}+O(n^{-2})-1\right) = \frac{p}{2\sqrt{n}} + O(n^{-3/2}) $$ and therefore $$ \lim_{n\to\infty}\big(a_{n+p}-a_n\big) = 0 . $$

Answer 2

Reason is, that in cauchy sequence definition de facto we have uniform convergence requirement with respect to $p$: $$\forall \varepsilon \gt 0, \exists N\in \mathbb{N}, \forall n \gt N, \forall p \in \mathbb{N}, |x_{n+p}-x_n|\lt \varepsilon$$

Here $N$ is independent from $p$.

We have one $N$ for all $p$.

If we move $\forall p \in \mathbb{N}$ from it's place after condition for $N$ before it, then we lost uniform dependence for $N$ from $p$ i.e. cauchy sequence property and obtain $$ \forall p \in \mathbb{N},\forall \varepsilon \gt 0, \exists N\in \mathbb{N}, \forall n \gt N, |x_{n+p}-x_n|\lt \varepsilon$$ Now $N$ is dependent also on $p$, while before is not so.

For each $p$ we have corresponding $N$.

Such sequence can be even divergent as is seen from example $$x_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$$ for which for every $p$ holds $$x_{n+p}-x_n=\frac{1}{n+1}+\cdots+\frac{1}{n+p}\leqslant \frac{p}{n+1}\to 0$$

Answer 3

example:

the increasing sequence $$a_n:=\log(n)$$ is not bounded but $$|a_{n+p}-a_n|=\log(n+p)-\log(n)=\log \frac{n+p}n=\log(1+\frac pn) \xrightarrow[n\to \infty]{}\\\lim_\limits{n\to \infty}\log(1+\frac pn)=\log(1+\lim_\limits{n\to \infty}\frac pn)=\log(1)=0\\$$ because the function $\log$ is continous at $1$.

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Here is the example already posted by @GEdgar presented in the way @ThomasAndrews proposes:

the increasing sequence $$a_n:=\sqrt{n}$$ is not bounded but $$|a_{n+p}-a_n|=\sqrt{n+p}-\sqrt n=\frac{(\sqrt{n+p}-\sqrt n)(\sqrt{n+p}+\sqrt n)}{\sqrt{n+p}+\sqrt n}=\frac p {\sqrt{n+p}+\sqrt n}\le\frac p{2\sqrt{n}}\to0$$

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Both are the specialization of the following example.

Is $f:(L,\infty) \mapsto \mathbb R$ a differentiable function and $\lim_\limits{x\to\infty}f'(x)=0$

and $$a_n:=f(n)$$

then for $\xi_{n,p} \in (n,n+p)$

$$|a_{n+p}-a_n|=|f(n+p)-f(n)|=|f'(\xi_{n,p})|(n+p)-n|\xrightarrow[n\to\infty]{}0\\$$

by the Mean Value Theorem.

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@zkutch proposes the follwoing example:

$$a_n:=\sum_{k=1}^{n}\frac 1 k$$ We get

$$|a_{n+p}-a_n|= \sum_{k=n}^{n+p}\frac 1 k \le \frac{p+1}n\xrightarrow[n\to\infty]{}0\\$$

This can be generalized in the following way:

$b_n$ is a sequence such that $b_n\to0$ and $a_n=\sum_{k=1}^{n}b_k$. Then for every $p$

$$|a_{n+p}-a_n|= \sum_{k=n}^{n+p}b_k=\sum_{k=0}^{p}b_{n+k}\xrightarrow[n\to\infty]{}\lim_{n\to\infty} \sum_{k=0}^{p}b_{n+k}=\sum_{k=0}^{p}\lim_{n\to\infty}b_{n+k}=\sum_{k=0}^{p}0=0$$

We know that the convergence of $\sum_{k=1}^na_k$ requires that $a_k$ is a null sequence, but that it is not sufficient for the existence of $\sum_{k=1}^na_k$ that $a_k$ is a null sequence. For all such null sequences $a_k$ where $\sum_{k=1}^na_k$ does not converge, $\lim_{n\to \infty}|a_{n+p}-a_n|= 0$ for all $p$.

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So we can construct examples from null sequences and from functions with derivates that tend o $0$ if $x$ tends to infinity.