Is this implication true?

Question

Suppose that a real sequence $u_n$ is such that $$u_{n+1}-u_n \rightarrow0$$

That is not enough to prove that $u_n$ is convergent (take $u_n=ln(n)$)

Now what if $u_n$ is bounded ? I guess it does converge, but how to prove this ? I tried to show that it had only one accumulation point...

It doesn't need to converge. Let $(u_n)$ be a sequence oscillating between $0$ and $1$ with $\lvert u_{n+1} - u_n\rvert = \frac{1}{n+1}$. Or, similar and explicit, $u_n = \sin \log (n+1)$. — Daniel Fischer, Oct 10 '13 at 20:03
Found it! it has been asked before http://math.stackexchange.com/questions/359344/give-an-example-of-a-bounded-non-convergent-real-sequence-a-n-s-t-a-n-a/359353#359353 — clark, Oct 10 '13 at 20:10
http://math.stackexchange.com/questions/121406/set-of-cluster-points-of-a-bounded-sequence is also interesting. — Philippe Malot, Oct 10 '13 at 20:33

score 5 · Answer 1 · answered Oct 10 '13 at 20:05

Bounded does not help. Use the sequence made up of partial sums from the sequence $$1, -\frac{1}{2},-\frac{1}{2}, \frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}, -\frac{1}{8}, -\frac{1}{8},-\frac{1}{8},-\frac{1}{8},-\frac{1}{8},-\frac{1}{8},-\frac{1}{8},-\frac{1}{8},\dots.$$

Antoine · Answer 2 · 2013-10-10T20:14:11.187

3

The sequence $$0, 1, \frac{1}{2}, 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1, \frac{7}{8},\frac{6}{8},\frac{5}{8}, ..., \frac{1}{8}, 0 , \frac{1}{16}, \frac{2}{16}, ..., 1 , \frac{31}{32}, ...$$

is not convergent and it has infinitely many accumulation points (every $x\in [0,1]$ is accumulation point).

edited Oct 10 '13 at 20:14

answered Oct 10 '13 at 20:08

Antoine

3,439

1

Is this André's sequence in disguise? :-) – robjohn Oct 10 '13 at 20:14
Indeed!:) (Almost, to be precise. If Andre had started with $0$ ...) But it was "discovered" independantely:) – Antoine Oct 10 '13 at 20:16

robjohn · Accepted Answer · 2013-10-10T20:20:45.917

3

Hint: Consider $$ u_n=\sin(\log(n)) $$ Show that $|u_{n+1}-u_n|\lt\frac1n$ yet $\limsup\limits_{n\to\infty}u_n=1$ and $\liminf\limits_{n\to\infty}u_n=-1$.

edited Oct 10 '13 at 20:20

answered Oct 10 '13 at 20:11

robjohn

345,667

score 3 · Answer 4 · answered Oct 10 '13 at 20:15

In $\mathbf R$ it won't be a sufficient condition. But if you want to see spaces where this condition is really useful, you can look about ultrametric spaces. It's the name for metric spaces $(X,d)$ where $\forall (x,y,z)\in X^3, d(x,y)\leq\sup(d(x,z),d(z,y))$ (an stronger inequality of triangle). In this spaces a sequence is a Cauchy sequence if and only if $d(u_n,u_{n+1})\rightarrow 0$. So in a complete ultrametric space, a sequence is convergent if and only if $d(u_n,u_{n+1})\rightarrow 0$.

I notice now, that in $\mathbf R$, the sequence $u_n$ is convergent if and only if the serie $\sum (u_n- u_{n-1})$ is convergent (for example if $u_n-u_{n-1}=O(1/n^a)$ where $a>1$.

$d$ is a distance on $X$ i.e a fonction from $X\times X$ to $R^+$ such as : -d is symmetric ($d(x,y)=d(y,x)$) -$\forall(x,y,z)\in X^3, $d(x,y)<d(x,z)+d(z,y)$ — D.L., Oct 10 '13 at 20:20
And in ultrametric spaces, you want also that $d$ verifies the ultra metric inequality : $\forall (x,y,z)\in X^3, d(x,y)<sup(d(x,z),d(y,z))$ — D.L., Oct 10 '13 at 20:23

Is this implication true?

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