On Complex sequences inequalities

Question

Attempt:

(a) If $m>n$, Notice that $\dfrac{1}{1+|m-n|} < \dfrac{1}{|m-n|} < \dfrac{1}{m-n} < \dfrac{1}{n} $. So that if choosing $N > 1/\epsilon$ then we obtain $|z_m-z_n| < \epsilon$ so (z_n) is cauchy so convergent. I claim that limit is $0$. We need to observe that as $n$ gets large, the separation of the terms of the sequence gets smaller. In fact, if $m = 2n $, then

$$ |z_{2n} - z_n | < \dfrac{1}{1+n} < \dfrac{1}{n} \to \infty $$

Im having some difficulties making this formal. Any suggestion?

As for (b), we know $|w_n| < B$ for all $n$ and after some $N$, we have that $|z_n| < \epsilon / B $. So, with this same $N$, we obtain that

$$ | w_n z_n | < B \cdot \frac{ \epsilon }{B} = \epsilon $$

and we have the result.

(c). We can find $N$ so that for $n>N$ we have $|z_n - A| < \epsilon $

So that with this same $N$, we have

$$ \left| \dfrac{ z_1 + ... + z_n }{n } - A \right| = \left| \dfrac{ z_1+...+z_n - n A }{n} \right| \leq \dfrac{ \sum_{i=1}^n |z_i - A | }{n} \leq \dfrac{ n \sup_{n \in \mathbb{N} } |z_n - A | }{n} < \epsilon$$

and result follow. Is this a valid argument?

Answer 1

The answer provided by Clement Yung for the first sub-question is correct, but can be amended a little. The condition $$|z_m-z_n|\leq \frac{1}{1+|m-n|}$$ is not just satisfied by constant sequences; it is satisfied only by constant sequences. Let $m$ and $n$ be arbitrary indices such that $d=|z_m-z_n|>0$ and set $k=\lfloor \max(m,n)+\frac{2}{d}\rfloor$. According to the given conditions, we have

$$|z_k-z_m| \leq \frac{1}{1 + |k-m|} < \frac{1}{2/d} = \frac{d}{2}$$ and the same applies to $z_n$ too. But that yields a contradiction: $$|z_m-z_n| = |z_m-z_k + z_k-z_n| \leq |z_m-z_k| + |z_k-z_n| < d$$

Answer 2

For (a), unless some information is missing, there's no way we can determine the limit of the sequence. Let ${z_n}$ be any constant sequence, and we have that: $$ |z_n - z_m| = 0 \leq \frac{1}{1 + |m - n|} $$ However, the limit of the constant sequence is clearly the constant itself.

For (b), your proof is good.

For (c), your attempt is incorrect. This is because if $\{z_n\}$ is not a constant sequence, say $z_k \neq A$, then: $$ \sup_{n \in \Bbb{N}} |z_n - A| \geq |z_k - A| $$ So we can't let $\epsilon$ be arbitrarily small in your working. Try referring to this.

Answer 3

For a) we can note that for any fixed $k$ we have $$ 0\leq|z_n-z_k|\leq \frac{1}{1+|n-k|}. $$ When $n\to\infty$ the right hans side tends to $0$, so $\lim\limits_{n\to\infty}|z_n-z_k|=0$. Hence, the limit $\lim\limits_{n\to\infty} z_n$ exists and is equal to $z_k$ for all $k$. Therefore, the sequence $\{z_n\}$ is constant.