- #1

fishturtle1

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## Homework Statement

Show that the sequence with two distinct accumulation points must diverge. (Hint: look at the proof of divergence for {##(-1)^k##}.

## Homework Equations

Some definitions and propositions I'm trying to use:

2.2.3: A sequence cannot converge to two different numbers. If {##x_k##} is a sequence in ##\mathbb{R}## which converges to both y and z, then y = z.

2.2.5 A real sequence {##x_k##} diverges to (positive) infinity (denoted ##x_k \rightarrow \infty##) if for every ##M \epsilon \mathbb{R}## there exists ##K## such that, for every ##k \ge K##, we have ##x_k \ge M##.

2.3.9 For any sequence {##x_k##} in ##\mathbb{R}##, an accumulation point of the sequence is any point which is the limit of some subsequence {##y_j##} of the sequence {##x_k##}.

The proof/excerpt for {##(-1)^n##} diverges is this:

While every convergent sequence must be bounded, it is not true that every bounded sequence vineries: in other words, there exist bounded sequence which diverge .For example, consider the sequence

$$\lbrace(-1)^n\rbrace = +1, -1, +1, -1, ...$$

which alternates the two values +1 and -1. This is bounded, since the absolute value of al elements is 1. Nonetheless the sequences diverges. To see this consider the possibilities. It can't converge to +1, because every second element satisfies

$$|(-1)^k - 1| = 2$$

so it cannot approximate +1 with accuracy, say ##\epsilon = 1##. However, if

##y \neq 1##, we see that every even index term has

$$|(-1)^k - y| = |y - 1| > 0$$

and so we can't approximate y with accuracy, say, half the distance from y to 1:

##\epsilon = |y-1|/2##.

## The Attempt at a Solution

Proof: Let {##x_k##} be a sequence and {##y_j##}, {##z_h##} be subsequences of {##x_k##} with distinct accumulation points. Then ##\lim_{j\to\infty} y_j = a## and ##\lim_{h \to\infty} z_h = b## where ##a \neq b##.

...

edit1: one of the things I'm stuck on is that, what if I have the sequence {##x_k##} = (-10, -5, 0, 5, 10, 5, 0, 0, 0, 0, 0). Then I looked at the subsequences {##y_j##} = (-10, -5, 0, 5, 10) and {##z_h##} = (5, 0, 0, 0). Then {##x_k##} converges to 0, {##y_j##} approaches 10 and {##z_h##} approaches 0. So what's the problem?

please help