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Today I learned about Cauchy Sequences, defined as the following:

A sequence $(x_n)$ is a Cauchy Sequence if $\forall\varepsilon>0,\exists N\in\mathbb{N}\ \forall n,m\geq N: |x_n-x_m|<\varepsilon$.

Assuming we are dealing with a complete metric space, all Cauchy Sequences converge, right? If so, for what sequences is it easier to show they are Cauchy in order to show they converge (as opposed to the limit definition of a convergent sequence)? My professor said they tend to be sequences whose limit is not immediately clear but I can't think of any such examples. Could someone please share an example of a convergent sequence whose limit and perhaps bounds are non-trivial but can be proven to be convergent by proving they are Cauchy?

Thank you!

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    $\begingroup$ Maybe $a_n=1-{1\over2}+{1\over3}-\cdots +(-1)^{n+1} {1\over n}$. It's not to hard too show $(a_n)$ is Cauchy. $\endgroup$ – David Mitra Nov 5 '19 at 5:12
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A family of such examples comes from the following theorem:

Theorem. Let $(x_n)$ be a sequence of real numbers. Suppose that $$\sum_{n=1}^{\infty} |x_{n+1} - x_{n}| < \infty. $$ Then $(x_n)$ converges.

The proof goes by showing that $(x_n)$ is a Cauchy sequence, and in fact, one can show that this statement is equivalent to Cauchy completeness. Now using this, one can cook up various example whose limit is by no means a known expression in terms of familiar functions and/or constants. For instance, choose an arbitrary assignment of signs $e_k\in\{-1,+1\}$ and set $x_n=\sum_{k=1}^{n} e_k/k^2$. Then the above theorem tells that $(x_n)$ converges, but except for a handful of examples, the exact nature of the limit is not known to us. And even in the case $e_k = 1$, the limit $\sum_{k=1}^{\infty} 1/k^2$ is given by the highly non-trivial value $\pi^2/6$, which is the content of the famous Basel Problem.

Another line of examples may be drawn from the following corollary of this theorem:

Contraction Mapping Theorem. Let $K\subseteq\mathbb{R}$ be closed and $k \in [0, 1)$. Suppose $f:K\to K$ satisfies $$ |f(x)-f(y)| \leq k|x -y|, \qquad \forall x, y \in K. $$ Then there exists a unique $c \in K$ such that $f(c) = c$ holds. Moreover, if $f^{\circ n}$ denotes the $n$-fold composition of $f$, then $$\lim_{n\to\infty} f^{\circ n}(x) = c, \qquad \forall x \in K. $$

The proof goes by showing that $x_n := f^{\circ n}(x)$ satisfies $|x_{n+1} - x_n| \leq k^n|x_1 - x_0|$ and thus the above theorem is applicable. Now, as a curious example, we consider the function $f : [0,1] \to [0, 1]$ defined by $f(x) = \cos(x)$. Mean Value Theorem tells that the above theorem holds with $k = \sin(1) \in [0, 1)$, and so, there exists a unique solution of the equation $f(c) = c$ on $[0, 1]$ and $f^{\circ n}(x) \to c$ as $n\to\infty$ for all $x \in [0, 1]$.

Plot of x and cos(x)

And again, we do not know how to write $c$ in 'closed form'.

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  • $\begingroup$ Very interesting! I hadn't even thought of sequences as being such abstract objects. The example at the end is also wonderful, thank you very much for your detailed answer! $\endgroup$ – Crystal Nov 6 '19 at 2:28

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