# Cauchy sequence and topological problems

1. Oct 16, 2011

### furi0n

[PLAIN]http://img805.imageshack.us/img805/1575/photo0138d.jpg [Broken]
Hi, on thursday, i have exam of advance calculus and i could not solve two problem in study sheet given by İnstructor. By 9 question, i prove by add an subtract XnY to |XnYn-XY| and i have found that |Xn(Yn-Y)+ Y(Xn-X)|<=|Xn(Yn-Y)|+|Y(Xn-X)|<=|Xn||Yn-Y)|+|Y||(Xn-X)| but, to use the last inequality, i need to know that Xn=a(Yn-Y) but how can i get that?

by last question, i could not have a idea and apply Cauchy properties to this :(..

Last edited by a moderator: May 5, 2017
2. Oct 16, 2011

### spamiam

For #9, it would be helpful to show that a convergent sequence is bounded. Can you prove that?

For #10, have you seen the monotone convergence theorem? If so, prove inductively that the sequence is monotone and bounded.

I'm not sure what your professor was hinting at, but I'll think about it more...

3. Oct 16, 2011

### HallsofIvy

Staff Emeritus
No, you don't. You just need that |Xn| is bounded and that follows from the fact that it converges.

Well, you are given a hint.
$$|x_{n+2}- x_{n+1}|= |\frac{1}{3+ x_{n+1}}- \frac{1}{3+ x_n}|= |\frac{3+x_n- (3+ x_{n+1}}{(3+ x_{n+1})(3+ x_n)}|= \frac{x_{n+1}- x_n}{(3+x_{n+1})(3+x_n}$$
But it is clear that all $x_n$ are positive so $(3+ x_{n+1})(3+ x_n)> 9$.

Now, with $x_1= 1$, $x_2= 1/(4)= 1/2$ so that $x_3- x_2\le (1/9)|x_2- x_1| 1/9(1/2)$, $|x_4- x_3|\le (1/9)|x_3- x_2|\le (1/2)(1/9)^2$, etc.

That's important because to use the Cauchy Criterion, we need to show that $|x_n- x_m|$ goes to 0 as m and n both go to infinity. We can, without loss of generality, assume that n>m and say $|x_n- x_m|\le |x_n- x_{n-1}|+ |x_{n-1}- x_m|\le |x_n- x_{n-1}|+ |x_{n-1}- x_{n-2}|+ |x_{n-2}+ x_m|$, etc., eventually arriving at
$|x_n- x_m|\le |x_n- x_{n-1}|+ \cdot\cdot\cdot+ |x_{m+1}- x_m|$
$|x_n- x_m|\le [(1/9)^m((1/2)(1/9)^{n-m}+ (1/2)(1/9)^{n-m-1}+ \cdot\cdot\cdot+ (1/2)]$
$|x_n- x_m|\le (1/2)(1/9)^m\sum_{i= 0}^{n- m} (1/9)^i$
which, because all terms are positive, is certainly less than the sum we get if we extend the sum to infinity:
$|x_n- x_m|\le (1/2)(1/9)^m\sum_{i=0}^{\infty}(1/9)^i$
which then is $(1/2)(1/9)^m/$ times a geometric series which can be summed for all n and then let m go to infinity also.

Once you have shown that this is a Cauchy sequence, and so converges, you take the limit on both sides of $x_{n+1}= 1/(3+ x_n)$ to find the limit.

Last edited by a moderator: May 5, 2017
4. Oct 17, 2011

### furi0n

@spamiam
for #9, it seems more complicated for me, and i think i can ask my classmates what is the relation between convergence and this question,
for #10, i know that theorem, but it does not monotone if i do correctly..