The Cauchy criterion is used to prove the convergence of sequences $(a_k)$ with unknown or irrational limit: If for every $\epsilon > 0$ there is a $k$ such that for $m, n > k$ we have $|a_n-a_m|<\epsilon$ then the sequence converges.
My question: What functions are best suited to show undergraduates that this criterion is useful? That means: I am looking for functions without an easily recognizable convergence that can easily be shown convergent by the Cauchy criterion.
Here are a couple of examples that I believe are fairly standard for illustrating the Cauchy convergence criterion.
Example 1: Let $a < b$ be real numbers and define the sequence $\{x_n\}$ of real numbers using $2$-term recursion as follows: $x_1 = a$ and $x_2 = b$ and, for each integer $n \geq 3,$ let $x_n = \frac{1}{2}(x_{n-2} + x_{x-1}).$ This sequence is not monotone, and thus methods using least upper bounds and greatest lower bounds are difficult to apply. For instance, the odd-numbered terms form an increasing sequence that is bounded above by $b$ (and thus converges), and the even-numbered terms form a decreasing sequence that is bounded below by $a$ (and thus converges), but how do we know these two subsequences converge to the same limit? However, it is easy to see that this is a Cauchy sequence --- for each integer $N \geq 2,$ if $m,n \, \geq \, N,$ then $x_m$ and $x_n$ both belong to the same $(N-2)$-bisected subinterval of $[a,b]$ and hence $|x_n – x_m| \, \leq \, \frac{1}{2^{N-2}}(b-a)$ --- and therefore this sequence converges. Incidentally, the limit of this sequence is $\frac{1}{3}(a+2b),$ which can be found by recasting the sequence as the partial sums of
$$a + (b-a) - \frac{1}{2}(b-a) + \frac{1}{4}(b-a) - \frac{1}{8}(b-a) + \cdots + (-1)^k2^{-k}(b-a) + \cdots $$
and noticing that this series is geometric beginning with the second term.
Example 2: We can show that the harmonic series $\sum\limits_{k=1}^{\infty}\frac{1}{k}$ diverges by showing that its sequence of partial sums is not a Cauchy sequence. For instance, letting $S_n = \sum\limits_{k=1}^{n}\frac{1}{k},$ we have
$$ S_{2n} \; - \; S_n \;\; = \;\; \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \;\; \geq \;\; n\left(\frac{1}{2n}\right) \;\; = \;\; \frac{1}{2} $$
One application is infinite series, since they are characterized by their sequences of partial sums. First, use the Cauchy Criterion for Sequences to establish a similar result for series. Then, use this criterion to more easily prove facts about series (e.g. Comparison Tests, Absolute Convergence Test, etc.) That is, for the students, you can frame the Cauchy Criterion as a way to make later proofs about series more direct and revealing. Rather than using the same technique to prove all those results, let's identify that common technique.
Theorem (Cauchy Criterion for Series): The series $\sum_{k=1}^\infty a_k$ converges if and only if, for any $\varepsilon>0$, there exists an $N\in \mathbb{N}$ such that, whenever $m>n\geq N$, it follows that $|a_{m+1}+a_{m+2}+\cdots +a_n|<\varepsilon$.
Proof: Notice that the partial sums $s_n=a_1+a_2+\cdots+a_n$ and $s_m=a_1+a_2+\cdots+a_m$ make $|s_n-s_m|=|a_{m+1}+a_{m+2}+\cdots +a_n|$. Apply the Cauchy Criterion for Sequences.
Theorem (Comparison Tests): Assumie $(a_k)$ and $(b_k)$ are sequences with $0\leq a_k\leq b_k$ for all $k$. Then:
These can be proven using the Cauchy Criterion for Series.
