# Question about Sequences - sorry if this is in the wrong place.

1. Jun 28, 2009

### AnthonyAcc

I'm taking an introductory class on analysis right now and I'm trying to get through the book that we are reading. I'm having difficulty understanding a park of it and was hoping someone could help me out. The part I'm reading about now is on null sequences:

Here's an excerpt. I'm having trouble entering in the math type so sorry for no symbols.

1. Every null sequence is a bounded sequence. For - choose epsilon=1 - we have the |Z_nu|<1 for nu > mu, and hence the |Z_nu| is less than or equal to K = max(1, |Z_0|, ... ,|Z_mu|).
2.Let {|Z_nu|} be a null sequence. Suppose that for a fixed K the terms of a sequence {Z'_nu} under investigation satisfy the condition that, for all nu after a certain stage mu',

|Z'_nu| is less than or equal to K|Z_nu|.

I have some trouble understanding what this means. Mu here is the stage beyond which |Z_nu| is less than epsilon. I dont really understand the part about K = max (etc) or the Z', mu' part towards the end.

Thanks.

2. Jun 28, 2009

### snipez90

Basically, K is the bound for the sequence. As you said it yourself, mu is the stage beyond which |Z_nu| is less than epsilon, which we chose as 1 (it could have been any positive number, but 1 is as good as anything). Thus, for any index after mu, the corresponding terms of the sequence are less than 1 (follows directly from the statement " For - choose epsilon=1 - we have the |Z_nu|<1 for nu > mu"). On the other hand, there are only finitely many terms before this index mu, i.e. for indices nu <= mu, there are only finitely many corresponding terms of the sequence. Thus, the maximum of absolute value of these terms, i.e. max{|Z_0|, ... , |Z_mu|} could also potentially serve as a bound for the entire sequence. Now considering all of the terms of the sequence together, we can choose K = max(1, |Z_0|, ... ,|Z_mu|) to be the bound on the sequence.

Last edited: Jun 28, 2009
3. Jun 28, 2009

"Every null sequence is a bounded sequence. For - choose epsilon=1 - we have the |Z_nu|<1 for nu > mu, and hence the |Z_nu| is less than or equal to K = max(1, |Z_0|, ... ,|Z_mu|)."

Recall that a null sequence converges to zero. In order to show that every null sequence is bounded you need to show that the absolute value of each term is smaller than constant. The point of your first bullet is that this can be done in two steps.

Step 1: Pick $$\varepsilon = 1$$. Because of the convergence we can find an integer N such that

$$|Z_n - 0|= |Z_n| < 1 \quad \forall n \ge N$$

Step 2: So now we know that infinitely many of the terms (not all of them) are bounded. What about the ones for $$m < N$$? Since there are finitely many of them, we can deal with them rather simply. Let

$$K = \max (1, |Z_0|, |Z_1|, \dots, |Z_{N-1}|)$$

Then, for any integer n, we are guaranteed to have

$$|Z_n | \le K$$

so every term in the sequence is bounded.

doesn't seem to be complete - is there something that follows? This simply states that you have two sequences, $$\{Z_\nu \}$$ and $$\{Z'_\nu \}$$ and, after some
point $$\mu$$, you have
$$|Z'_\nu| < |Z_\nu| \quad \forall \nu > \mu$$