- #26

- 144

- 12

The sequence ##{-(1)^n}_1^\infty## is -1, 1, -1, 1, -1, ...

Since the limit of the sequence can't found by inspection, so I suspect that as n approaching infinity, the limit of this sequence is 1 or -1 (by method number 3).

First, if

$$\lim_{n\to \infty} (-1)^n = 1$$

then for any small positive number ##\epsilon## there is natural number ##N## that makes ##|(-1)^n -1|<\epsilon##. If there is no natural number N, then 1 is not the limit of this sequence as n approaching infinity.

Proof:

$$|(-1)^n - 1|<\epsilon$$

$$(-1)^n <\epsilon + 1$$

Because ##\epsilon## is positive number, so ##\epsilon + 1## also a positive number (must bigger than zero). Is there any n satisfy this condition? Only the even number n that satisfy this condition. So, 1 is not the limit of this sequence.

Second, if $$\lim_{n\to \infty} (-1)^n = -1$$

then for any small positive number ##\epsilon## there is natural number ##N## that makes ##|(-1)^n + 1|<\epsilon##. If there is no natural number N, then -1 is not the limit of this sequence as n approaching infinity.

Proof:

$$|(-1)^n + 1|<\epsilon$$

$$(-1)^n <\epsilon - 1$$

Because ##\epsilon## is positive number, so ##\epsilon - 1## is a negative number (smaller than zero). Is there any n satisfy this condition? Only the odd number n that satisfy this condition. So, -1 is not the limit of this sequence.

From these 2 proofs, I can conclude that $$\lim_{n\to \infty} (-1)^n DNE$$

Now, I will try to use the more general method: number 1.

$$\lim_{n\to \infty} (-1)^n = L$$

$$|(-1)^n - L|<\epsilon$$

$$(-1)^n <\epsilon + L$$

If ##L > 0##, then only the even number of n satisfy that condition.

If ##L < 0##, then only the odd number of n satisfy that condition.

So, the limit is DNE.

Hmm. Are my works true? Thankss

Oh, I have read about Cauchy sequence, but need more explanation to identify whether a sequence is Cauchy sequence or not. May be some of you could give an example or explanation? Thanks again.