Do Cauchy sequences always converge?

  • #1
Calabi
140
2
Hello evry body let be $(u_{n}) \in \mathbb{C}^{\matbb{N}}$
with $u_{n}^{2} \rightarrow 1$ and $\forall n \in \mathbb{N} (u_{n+1) - u_{n}) < 1$.
Why does this sequences converge please?

Thank you in advance and have a nice afternoon:oldbiggrin:.
 
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  • #2
Calabi said:
Hello evry body let be $(u_{n}) \in \mathbb{C}^{\matbb{N}}$
with $u_{n}^{2} \rightarrow 1$ and $\forall n \in \mathbb{N} (u_{n+1) - u_{n}) < 1$.
Why does this sequences converge please?

Thank you in advance and have a nice afternoon:oldbiggrin:.

Why don't you fix the Latex?
 
  • #3
PeroK said:
Why don't you fix the Latex?
Plus consider please whether the difference ##(u_{n+1} - u_{n}) < 1## shouldn't be better ##|u_{n+1} - u_{n}| < 1##
 
  • #4
It would be better indeed. And how to put the latex here please?

Thank you in advance and have a good afternoon:oldbiggrin:.
 
  • #5
Calabi said:
It would be better indeed. And how to put the latex here please?

Thank you in advance and have a good afternoon:oldbiggrin:.

You double your dollars!
 
  • #6
PeroK said:
You double your dollars!
If it was that easy ...
 
Last edited:
  • #7
So did anyone has a solution please?
 
  • #8
Ok, I fixed Calabi's TeX-notation. See and learn...

"Hello evry body let be ##(u_{n}) \in \Bbb{C}^{\Bbb{N}}##
with ##u_{n}^{2} \rightarrow 1## and ##\forall n \in \Bbb{N}\,(u_{n+1}-u_{n})<1##.
Why does this sequences converge please?"

For large ##n##, ##u_n^2## is close to ##1##. What can you then say about ##u_n## for large ##n##? Can the sign of ##u_n## alter for large ##n##, considering that ##u_{n+1}-u_n<1##?
 
  • #9
Calabi said:
So did anyone has a solution please?
It's good to see that the quality of the discussion in the technical math sections has reached a new high.
 
  • #10
i know what you mean but, in general, complex numbers have no "sign", so your nice hint needs a modified formulation.
 
  • #11
Hello every one sorry Kyrlov I should give what I search on this exercicse even if it's wrong,
so as $$u_{n}^{2} \rightarrow 1$$ we can say that $$\forall \epsilon > 0, \exists N \in \mathbb{N} / n > N \Rightarrow |u_{n}^{2} - 1| < \epsilon$$. As Erland said it could be a story of sign but we're in $$\mathbb{C}$$.
 
  • #12
Ah, OK, but in that case, the condition ##u_{n+1}-u_n<1## is meaningless, since order does not exist between complex numbers. Presumably, Calabi meant ##|u_{n+1}-u_n|<1##. If so, let us instead ask if the sign of the real part of ##u_n## can alter for large ##n##...
 
  • #13
Are the ##u_n##:s close to some particular set of numbers for large ##n##?
 
  • #14
Ther's 2 possible of numbers -1 or 1. Like the square of u converge in 1 and as $$|u_{n+1} - u_{n}| < 1$$ the term of $$u_{n}$$
are in one neighbourhood of 1 or -1 I see it but I can't say it good.
 
  • #15
Yes, if we choose small neighborhoods of ##1## and ##-1##, all ##u_n## is in one of them, not the other, for all sufficiently large ##n##, since otherwise we would have ##|u_{n+1}-u_n|\ge 1## for some ##n##. This holds for arbitrarily small neighborhoods of ##1## or ##-1##, so the sequence converges to ##1## or ##-1##. It just remains to make this argument formal.
 
  • #16
Yeah but it's this formalism I can't wright.
 
  • #17
Calabi said:
Yeah but it's this formalism I can't wright.
Try! How would you begin?
 
  • #18
We can do it in the absurd way : let suppose for exemple that $$\exists \epsilon > 0/ \forall N \in \matbb{N} \exists n > N / |u_{n} - 1| > \epsilon (1)$$. For this $$\epsilon > 0, \exists N \in \mathbb{N} / n > N \Rightarow |u_{n}^{2} - 1| < \epsilon $$.
For this N we can find n as in (1). For this n $$|u_{n+1} - u_{n}| < 1$$.
Hum.
I don't see the absurdity.
 
  • #19
Hmm, this might lead somewhere. What if you also consider the corresponding relation as your first one but for -1 instead of 1? Also, what if the ##\epsilon##:s you use are unequal, but perhaps related in some way?
 
  • #20
Calabi said:
Hello every one sorry Kyrlov I should give what I search on this exercicse even if it's wrong,
so as $$u_{n}^{2} \rightarrow 1$$ we can say that $$\forall \epsilon > 0, \exists N \in \mathbb{N} / n > N \Rightarrow |u_{n}^{2} - 1| < \epsilon$$. As Erland said it could be a story of sign but we're in $$\mathbb{C}$$.

Can you see why ##u_n## converges? If not, you'll never prove it using ##\epsilon##.

I'd draw a diagram for ##u_n^2## and try to see geometrically why ##u_n## must converge. Once you understand that, you can try to translate it into a formal ##\epsilon## proof.
 
  • #21
I made a proof : I whright it in french but the idea : for a big N I've got forall n > N $$\frac{1}{\sqrt{2}} < |u_{n}| < \sqrt{\frac{3}{2}$$. Then by recurrence I show that all the $$u_{n}$$ have a constant sign by using the hypothesis that $$|u_{n+1} - u_{n}| < 1$$.
So as $$\sqrt{u_{n}^{2}} = |u_{n}|$$ twoo case : if u is positive it convegre in 1 otherwise in -1.
That was for $$\mathbb{R}$$.
In $$\mathbb{C}$$ I can do a similar geometric prof.
 
  • #22
Calabi said:
I made a proof : I whright it in french but the idea : for a big N I've got forall n > N $$\frac{1}{\sqrt{2}} < |u_{n}| < \sqrt{\frac{3}{2}}$$. Then by recurrence I show that all the $$u_{n}$$ have a constant sign by using the hypothesis that $$|u_{n+1} - u_{n}| < 1$$.
So as $$\sqrt{u_{n}^{2}} = |u_{n}|$$ twoo case : if u is positive it convegre in 1 otherwise in -1.
That was for $$\mathbb{R}$$.
In $$\mathbb{C}$$ I can do a similar geometric prof.

Yes, you can use the same idea for complex numbers.

Hint: for complex numbers ##|u_n^2 - 1| = |u_n - 1||u_n + 1|##
 
  • #23
PeroK said:
Hint: for complex numbers |u2n−1|=|un−1||un+1|
And what you want to do with that please?
 
  • #24
Calabi said:
And what you want to do with that please?

That's the algebraic key to this problem.
 

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