Pseudo-Cauchy sequence

Take $$a_n = \sum_{k=1}^n {1\over k}.$$


Consider the sequence $$0,1,\frac{1}{2},0,\frac{1}{3},\frac{2}{3},1,\frac{3}{4},\frac{2}{4},\frac{1}{4},0,\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},1,\frac{5}{6},\frac{4}{6},\frac{3}{6},\frac{2}{6},\frac{1}{6},0, \frac{1}{7},\frac{2}{7},\cdots.$$ This is a pseudo-Cauchy sequence, and every real number between $0$ and $1$ is the limit of a subsequence of our sequence.


For the record it doesn't even work if you ask $a_n$ to be bounded. Take $a_n = \sin(\sum_1^n 1/k)$. Since $\sum_1^n 1/k → ∞ $ and also the difference $\left|\sum_1^{n+1} 1/k - \sum_1^n 1/k\right| = \frac{1}{n+1} \to 0$, $a_n$ oscillates between 1 and -1 forever. But as $\sin$ is Lipschitz, $$|a_{n+1} - a_{n}| \leq \left|\sum_1^{n+1} \frac{1}{k} - \sum_1^n \frac{1}{k} \right| = \frac{1}{n+1} → 0$$

Tags:

Real Analysis

Cauchy Sequences

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