2
$\begingroup$

$\textbf{Question}$: Let $\{z_n\}_{n=1}^{\infty}$ be any complex series while $\sum\limits_{n=1}^{\infty}\left|z_{n+1}^{-1}-z_n^{-1}\right|=\infty$, suppose $\sum\limits_{n=1}^{\infty}a_n z_n$ is convergent.$a_{n}$ is real.

Prove that: $$ \lim\limits_{N\rightarrow\infty}\left(\sum\limits_{n=1}^N a_n\right)\left(\sum\limits_{n=1}^N\left|z_{n+1}^{-1}-z_n^{-1}\right|\right)^{-1}=0. $$

$\textbf{Hint}$

(1) May use $\textbf{Summation by Parts}$

(2) May use the $\textbf{Lemma}$

Suppose $\varphi(n) \searrow 0(n \to \infty)$, and $\sum\limits_{n=1}^{\infty}a_{n}\varphi(n)$ is convergent,so we have: $$ \lim_{n \to \infty} \left( a_{1}+a_{2}+\cdots+a_{n} \right)\varphi(n)=0$$

$\textbf{My attempts and questions}$

First if $z_{n}$ is a constant sequence , so the result is trivial.

While $z_{n}$ is not a constant sequence. First by summation by parts I get $$ \left| \sum\limits_{n=1}^{N}a_{n}\right| = \left| \sum\limits_{n=1}^{N}a_{n}z_{n}z_{n}^{-1}\right| \leq M \left\{ \sum\limits_{n=1}^{N-1}\left|z_{n+1}^{-1}-z_{n}^{-1}\right| + \left|z_{N}^{-1}\right| \right\}, $$ where M = $\max\left\{ |a_{1}|,|a_{1}+a_{2}|,\cdots,|a_{1}+a_{2}+\cdots a_{N}| \right\}$

Next $$\left(\sum\limits_{n=1}^N a_n\right)\left(\sum\limits_{n=1}^N\left|z_{n+1}^{-1}-z_n^{-1}\right|\right)^{-1} \leq \dfrac{M \left\{ \sum\limits_{n=1}^{N-1}\left|z_{n+1}^{-1}-z_{n}^{-1}\right| + \left|z_{N}^{-1}\right| \right\}}{\sum\limits_{n=1}^N\left|z_{n+1}^{-1}-z_n^{-1}\right|} \leq M + \dfrac{M \left|z_{N}^{-1}\right| }{\sum\limits_{n=1}^N\left|z_{n+1}^{-1}-z_n^{-1}\right|}$$ Then I didn't know what to do next, I don't know the limitation of the rightmost term in the above equation, and I don't know how to estimate it.

$\endgroup$
2
  • $\begingroup$ The $a_n$ are real? $\endgroup$ Commented Sep 6, 2024 at 9:17
  • $\begingroup$ For convenience, assume $a_{n}$ is real $\endgroup$ Commented Sep 6, 2024 at 9:35

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.