Give an example of a bounded, non-convergent real sequence $(a_n)$ s.t. $a_n-a_{n-1}\rightarrow 0$

Question

This was the last part of an exam question. I have spent a long time looking for such an example but have failed (sequences when one of the three conditions is dropped are easily found but I couldn't find one when all three are required). Could you also give the way you thought of the example in your answer please?

Thanks!

$a_n - a_{n-1} + a_{n-1} - a_{n-2} \to 0$, iterating isn't it true that $a_n - a_1 \to 0$? — , Apr 12 '13 at 11:21
Take some divergent series $\sum_{i=1}^{\infty}x_n$ with $0<x_n<1$ and $x_n\to 0$. Then define $a_n$ to follow this series but change direction every time your partial sums near $1$ or $-1$. — Karl Kroningfeld, Apr 12 '13 at 11:34
See also http://math.stackexchange.com/questions/1437381/bounded-sequence-that-converges-to-zero — Martin Sleziak, Sep 16 '15 at 06:24

Hagen von Eitzen · Accepted Answer · 2013-04-12T11:30:36.847

6

You can verify that $a_n=\sin\ln n$ does what you want.

How did I find that example? The sine makes sure that the values remain bounded in $[-1,1]$. The logarithm grows unbounded so that the sequence will keep wandering between $-1$ and $+1$ (and not stay near one value). The growth of the logarithm gets slower and slower for larger $n$, which keeps $a_n-a_{n-1}$ small (note that $|a_n-a_{n+1}|= \frac d{dx}\sin\ln x=\frac1x\cos\ln x$ for some $x\in(n-1,n)$, hence $|a_n-a_{n+1}| <\frac1{n-1}$).

edited Apr 12 '13 at 11:30

answered Apr 12 '13 at 11:24

Hagen von Eitzen

382,479

Thanks this is very helpful. – Apr 12 '13 at 11:38

score 5 · Answer 2 · answered Apr 12 '13 at 11:24

5

Try $a_n=\sin(\log n)$ for every $n\geqslant1$. Or $a_n=\sin(\sqrt{n})$. Or $a_n=\sin(b_n)$ for any sequence $(b_n)$ such that $b_n\to\infty$ and $|b_{n+1}-b_{n}|\to0$.

answered Apr 12 '13 at 11:24

Did

284,245

You beat me - or rather my first answer attempt was too short :) – Hagen von Eitzen Apr 12 '13 at 11:25
Is it a standard example? I mean I don't think I could think of that! – Apr 12 '13 at 11:26
@HagenvonEitzen Sorry... – Did Apr 12 '13 at 11:26
@HagenvonEitzen And I was just about to post exactly the same answer as well. Surely this cannot be the only example...? ;-) – WimC Apr 12 '13 at 11:27
@DaochenWang Yes, see "expanded" version which (1) gives a host of other examples and (2) explains where they are all coming from. – Did Apr 12 '13 at 11:27
Someone just pointed that this answer is similar to this one (+1). Besides, you were first. – robjohn Oct 11 '13 at 03:18

Give an example of a bounded, non-convergent real sequence $(a_n)$ s.t. $a_n-a_{n-1}\rightarrow 0$

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