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This seems an elementary problem, but I don't know of any reference to it in the literature.

Consider the sequence $(a_n)_{n=1}^\infty$ of real numbers. Suppose $|a_{n+1}-a_n|\rightarrow0~(n\rightarrow\infty)$, and suppose furthermore that $|a_n|\nrightarrow\infty$. May we conclude that $a_n$ converges?

The second condition precludes the standard example $a_n=\sum_{j=1}^nj^{-1}$, which obviously tends to $+\infty$.

Stromael
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    Just ping-pong a sequence between e.g. $1$ and $-1$ with steps of length $\frac{1}{n}$. – Daniel Fischer Jul 01 '15 at 12:05
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    @DanielFischer Simple but nice idea, +1. I found similar solution and posted it before I realised it's basically the same what you wrote before. – CiaPan Jul 02 '15 at 00:11
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    It doesn't take much to change this into the definition of a Cauchy sequence; you just need $|a_m-a_n|\to0$ as $(m,n)\to\infty$. – Akiva Weinberger Feb 02 '18 at 16:25
  • Here is a counter example: $a_n=\frac{1}{n}((-1)^{n})$ – makansij Feb 27 '19 at 06:36
  • @guimpton, your $a_n$ converges to $0$, making it not a counter-example. If $\varepsilon>0$ take $N>1/\varepsilon$, so that $|a_n| = 1/n <\varepsilon$ for every $n>N$. – Stromael Feb 28 '19 at 11:09

4 Answers4

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No, we may not conclude that $a_n$ converges.

Consider

$$a_n=\sum_{j=1}^ns_jj^{-1}$$

where $s_j=\pm 1$. We can choose the $s_j$ so that they are positive until $a_n$ becomes greater than one, then negative until $a_n$ becomes less than zero, then positive until $a_n$ becomes greater than one, and so on. Then $a_n$ does not approach infinity but it never settles down to any limit, oscillating between zero and one.

Rory Daulton
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$a_i = \sin(\sqrt i)$ oscillates 'forever', never stabilizing, so it is obviously not convergent, however oscillations slow down as $i$ grows: $\sqrt{i+1}-\sqrt i$ decreases with $i$ growing, so does a sine value change $\operatorname{abs}\left(\sin(\sqrt i) - \sin(\sqrt {i-1})\right)$.

CiaPan
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The sequence won't necessarily converge.

For instance we can define the following sequence :

Let $i_0=1$, $i_1$ be the smallest integer such that such that $\displaystyle\sum_{k=i_0+1}^{i_1}\frac{1}k>1$, $i_2$ be the smallest integer such that such that $\displaystyle\sum_{k=i_1+1}^{i_2}\frac{1}k>1$ and so on, and let $f$ be the function who associates each $k\in[i_n,i_{n+1})$ with $i_n$.

Then $\displaystyle\sum_{k\in\mathbb{N}^*}\frac{1}k(-1)^{f(k)}$ diverges but satisfies your condition.

Hippalectryon
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The answer is negative consider $(a_n)$ defined for $2^{2m} \le n \le 2^{2m+1}-1$ by $a_n=\frac{k-2^{2m}}{2^{2m+1}-2^{2m}}$ and for $2^{2m+1} \le n \le 2^{2m+2}-1$ by $a_n=1-\frac{k-2^{2m+1}}{2^{2m+2}-2^{2m+1}}$

On top of your requirements, $(a_n)$ is having the full segment $[0,1]$ as limit points!

mathcounterexamples.net
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