# Cauchy sequences or not

1. Dec 7, 2015

### Calabi

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.

2. Dec 7, 2015

### PeroK

Why don't you fix the Latex?

3. Dec 7, 2015

### Staff: Mentor

Plus consider please whether the difference $(u_{n+1} - u_{n}) < 1$ shouldn't be better $|u_{n+1} - u_{n}| < 1$

4. Dec 7, 2015

### Calabi

It would be better indeed. And how to put the latex here please?

Thank you in advance and have a good afternoon.

5. Dec 7, 2015

6. Dec 7, 2015

### Staff: Mentor

If it was that easy ....

Last edited: Dec 7, 2015
7. Dec 10, 2015

### Calabi

So did anyone has a solution please?

8. Dec 10, 2015

### Erland

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. Dec 10, 2015

### Krylov

It's good to see that the quality of the discussion in the technical math sections has reached a new high.

10. Dec 10, 2015

### mathwonk

i know what you mean but, in general, complex numbers have no "sign", so your nice hint needs a modified formulation.

11. Dec 11, 2015

### Calabi

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. Dec 11, 2015

### Erland

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. Dec 11, 2015

### Erland

Are the $u_n$:s close to some particular set of numbers for large $n$?

14. Dec 11, 2015

### Calabi

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. Dec 11, 2015

### Erland

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. Dec 11, 2015

### Calabi

Yeah but it's this formalism I can't wright.

17. Dec 11, 2015

### Erland

Try! How would you begin?

18. Dec 11, 2015

### Calabi

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. Dec 11, 2015

### Erland

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. Dec 11, 2015

### PeroK

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. Dec 15, 2015

### Calabi

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. Dec 15, 2015

### PeroK

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. Dec 15, 2015

### Calabi

And what you wanna do with that please?

24. Dec 15, 2015

### PeroK

That's the algebraic key to this problem.