look-alike Cauchy sequence

Question

If $d(x_{n+1},x_n)<\frac{1}{n+1}$ then the sequence $\{x_n\}$ is a Cauchy sequence OR not? Here in this problem, why can I not let $n\to \infty$? Also for $d(x_{n+1},x_n)<\frac{1}{2^n}$, the sequence $\{x_n\}$ is given to be Cauchy. However, both seems Cauchy to me. How are they different?

Answer 1

The difference is, while $\frac{1}{n+1}$ is convergent, $\frac{1}{2^n}$ is summable, and that's the relevant piece of information.

Recall the definition of $x_n$ being Cauchy: for all $\varepsilon > 0$, there exists some $N$ such that $$m, n \ge N \implies d(x_n, x_m) < \varepsilon.$$ It needs more than consecutive terms $x_n, x_{n+1}$ to be close; we need all the tail terms to be close to each other.

By forcing $d(x_n, x_{n+1}) \le \frac{1}{n+1}$, we force consecutive terms to limit to $0$ distance, but we can still have sizable distances between other terms in the tail. Our best estimate for $d(x_n, x_{n+k})$ is \begin{align*} d(x_n, x_{n+k}) &\le d(x_n, x_{n+1}) + d(x_{n+1}, x_{n+2}) + \ldots + d(x_{n+k-1}, x_{n+k}) \\ &\le \frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{n+k} \\ &= \sum_{i=1}^{n+k} \frac{1}{i} - \sum_{i=1}^{n} \frac{1}{i} \end{align*} How do I know that this is our best estimate? It's not hard to construct an explicit real sequence for which this bound is sharp, i.e. where there are equalities at every step.

Note that this bound is, itself, a difference of partial sums of the (divergent) harmonic series. Since the sum is divergent, the partial sums cannot be Cauchy, hence such differences of partial sums can be made as large as you like. That is, there's no way to ensure $d(x_n, x_{n+k})$ is small.

(Of course, it's still possible that a given sequence satisfying $d(x_n, x_{n+1}) \le (n + 1)^{-1}$ is Cauchy, but by finding an example where our bound is sharp gives us a counterexample, proving that satisfying the inequality is not sufficient to be Cauchy.)

Compare and contrast this method with $2^{-n}$. Then the sum of the terms converges. Similarly, $$d(x_n, x_{n+k}) \le \sum_{i=1}^{n+k-1} \frac{1}{2^i} - \sum_{i=1}^{n-1} \frac{1}{2^i}.$$ Since the partial sums converge and hence are Cauchy, this bound can be made as small as we like, which is enough to show Cauchiness.