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up vote 3 down vote favorite

The following is the problem that I am working on.

Let $\{z_n\} = \{x_n\}+\{y_n\}$ be a sequence where $\{x_n\}$ is monotonically increasing, $\{y_n\}$ monotonically decreasing, and $\{z_n\}$ is bounded.

Is $\{z_n\}$ convergent ? What if $\{x_n\}$ and $\{y_n\}$ are also bounded ?

I can clearly see and prove that in the second case, $\{z_n\}$ must converge to the sum of each of the limits of the sequences (because they exist).

However, this is what I think about the case where only $\{z_n\}$ is bounded.

Intuitively I want to say that it would be nice if $\{z_n\}$ converges but that sounds too good.

So I considered the following case.

If $\{x_n\}$ increases "faster" than $\{y_n\}$ decreasing, $\{z_n\}$ will become a monotonically increasing sequence that is bounded, thus it will converge to the sup of $\{z_n\}$.

If $\{y_n\}$ decreases "faster" then with the similar argument $\{z_n\}$ will converge to the inf.

If their increasing and decreasing rate are equal then it's a constant sequence, so the limit is cogent.

But I am not 100% confident that there doesn't exist a "rate of increase and decrease that lies in between" so that $\{z_n\}$ eventually oscillates or something.

Can someone help me out ?

share|cite|improve this question
    
It is pretty easy to construct a sequence where $z_i = (-1)^i$, which is bounded but doesn't converge. –  Calvin Lin Jul 1 '13 at 1:50
    
@CalvinLin: From the answers I can see that, thank you. Do you have any more examples so that in the future I can get the idea of what kind of thinking I should be doing ? –  hyg17 Jul 1 '13 at 2:30
    
Generally, you should know why a counter-example works, and what 'property' it is exploiting. In this case, we want a bounded sequence that is not convergent, and the typical example is $(-1)^i$ (and equivalently $0,1,0,1,0,1, \ldots$). Then, you can think about how to create such a sequence $z_i$. Of course, it is not guaranteed that this approach always works, so understanding more/various counter-examples will be very helpful. –  Calvin Lin Jul 1 '13 at 2:33
    
You had a good analysis of what happens if $(x_n)$ increases faster than $(y_n)$ decreases, and what happens in the opposite case. All that remained is to realize that, since these are infinite sequences they can switch between the two cases. $(x_n)$ can increase faster than $(y_n)$ decreases for a while, say for $n\leq100$, and then it can be slower for a while, say for $100\leq n\leq200$, and then faster for a while, etc. That way you can manipulate $(z_n)$ to do just about anything you can imagine. –  Andreas Blass Jul 1 '13 at 3:02

2 Answers 2

up vote 7 down vote accepted

What about something like $ x_n = n + .5 \sin (n) $ and $ y_n = -n + .5 \sin(n) $?

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Wow, this is exactly the reason I was not confident with my argument. Is this a rather common technique ? –  hyg17 Jul 1 '13 at 2:32
up vote 4 down vote

Hint: $x_n = \lfloor \frac{n}{2} \rfloor$ and $y_n = - \lfloor \frac{n+1}{2} \rfloor$

share|cite|improve this answer
    
Ohhhhhh, that is clever ! I am not very articulate with the use of floor and roof functions, but this gives me great insight. Thanks ! –  hyg17 Jul 1 '13 at 2:35
    
@hyg17 Note that you could simply have created the sequences explicitly as $x_n = 0,0,1,1, 2, 2, 3, 3 \ldots $ and $y_n = -0, -1, -1, -2, -2, -3, -3, \ldots$. It's clear how to continue these sequences, and why it works. The use of floor/ceiling functions merely give a closed form for the sequence, which is not completely necessary. –  Calvin Lin Jul 1 '13 at 2:46
    
Ahh, thanks for loosening up some knots in my head. We are virtually at the same age but our brains are at a different level. lol –  hyg17 Jul 1 '13 at 2:53

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