§ An example of a sequence whose successive terms get closer together but isn't Cauchy (does not converge)


§ The problem

Provide an example of a sequence an:N→Ra_n: \mathbb N \rightarrow \mathbb R such that lim⁡n→∞∣an+1−an∣→0\lim_{n \rightarrow \infty} \vert a_{n+1} - a_n \vert \rightarrow 0, but lim⁡n,m→∞,m>n∣an−am∣≠0\lim_{n, m \rightarrow \infty, m > n} |a_n - a_m| \neq 0. That is, proide a series where the distances between successive terms converges to zero, but where distances between terms that are "farther apart than 1" does not converge to 0. That is, the sequence is not Cauchy .

§ Regular solution: Harmonic numbers


The usual solution is to take the harmonic numbers, Hn≡∑i=1n1/iH_n \equiv \sum_{i=1}^n 1/i. Then, we show that:
lim⁡n→∞∣Hn+1−Hn∣=∣1n+1−1n∣=1(n+1)n→0 \begin{aligned} \lim_{n \rightarrow \infty} \left| H_{n+1} - H_n \right| &= \left| \frac{1}{n+1} - \frac{1}{n} \right| \\ &= \frac{1}{(n+1)n} \rightarrow 0 \end{aligned}

lim⁡n→∞∣H2n−Hn∣=∣12n−1n∣=∑i=n+12n1n+1+1n+2+⋯+12n≥∑i=n+12n12n+12n+⋯+12n≥n2n=12≠0onx→∞ \begin{aligned} &\lim_{n \rightarrow \infty} \left| H_{2n} - H_n \right| \\ &= \left|\frac{1}{2n} - \frac{1}{n} \right| \\ &= \sum_{i=n+1}^{2n} \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} \\ &\geq \sum_{i=n+1}^{2n} \frac{1}{2n} + \frac{1}{2n} + \dots + \frac{1}{2n} \\ &\geq \frac{n}{2n} = \frac{1}{2} \neq 0 \text{on} x \rightarrow \infty \end{aligned}

§ Memorable solution: logarithm


We can much more simply choose an=log⁡(n)a_n = \log(n). This yields the simple calculation:
lim⁡n→∞an+1−an=log⁡(n+1)−log⁡(n)=log⁡((n+1)/n))=log⁡(1+1/n)→n→∞log⁡(1)=0 \begin{aligned} &\lim_{n \rightarrow \infty} a_{n+1} - a_n = \log(n+1) - \log(n) \\ &= \log((n+1)/n)) \\ &= \log(1 + 1/n) \xrightarrow{n \rightarrow \infty} \log(1) = 0 \end{aligned}

while on the other hand,
lim⁡n→∞a2n−an=log⁡(2n)−log⁡(n)=log⁡(2)+log⁡(n)−log⁡(n)=log⁡2≠0 \begin{aligned} \lim_{n \rightarrow \infty} a_{2n} - a_n = \log(2n) - \log(n) = \log(2) + \log(n) - \log(n) = \log 2 \neq 0 \end{aligned}

I find this far cleaner conceptually, since it's "obvious" to everyone that an=log⁡(n)a_n = \log(n) diverges, while the corresponding fact for HnH_n is hardly convincing. We also get straightforward equalities everywhere, instead of inequalities.
I still feel that I don't grok what precisely fails here, in that, my intuition still feels that the local condition ought to imply the Cauchy condition: if ana_n tells an+1a_{n+1} to not be too far, and an+1a_{n+1} tells an+2a_{n+2}, surely this must be transitive?
I have taught my instincts to not trust my instincts on analysis, which is a shitty solution :) I hope to internalize this someday.
EDIT: I feel I now understand what's precisely happening after ruminating a bit.
The Cauchy convergence criterion allows us to drop a finite number of terms, and then capture everything after that point in a ball of radius ϵ\epsilon. As ϵ\epsilon shrinks, all the terms in the sequence are "squeezed togeher".
In the an+1−ana_{n+1} - a_n case, only successive terms must maintain an ϵ\epsilon distance. But as the log⁡\log example shows, you can steadily plod along, keeping ϵ\epsilon ball next to ϵ\epsilon ball, to reach:
lim⁡n→∞lim⁡ϵ→0f(n)⋅ϵ \lim_{n \rightarrow \infty} \lim_{\epsilon \rightarrow 0} f(n) \cdot \epsilon
whose behaviour can do unexpected things depending on the choice of nn.