I need toSolve this proof for complex numbers

Question

Proof that If $$ \sum_{n=1}^{\infty} z_{n} $$ converges, then $$\lim _{n \rightarrow \infty} z_{n}=0 $$ Prove for complex numbers

Given an infinite sequence $z_{1}, z_{2}, z_{3}, \ldots ., $ let $$ S_{N}=z_{1}+z_{2}+\cdots+z_{N} $$ We call the sequence $ S_{1}, S_{2}, S_{3}, \ldots $ an infinite series, which we denote $$ \sum_{n=1}^{\infty} z_{n} $$ We call $ S_{N} $ a partial sum. If $ S_{N} $ converges with $S=\lim _{N \rightarrow \infty} S_{N}, $then we say $\sum_{n=1}^{\infty} z_{n} $ converges and write $$ \sum_{n=1}^{\infty} z_{n}=S $$ If $S_{N} $ does not converge, we say $\sum_{n=1}^{\infty} z_{n}$ diverges. Suppose $z_{n}=x_{n}+i y_{n} $ and $S=X+i Y . $ Then it follows from previous results that $$ \sum_{n=1}^{\infty} z_{n}=S $$ and only if $$ \sum_{n=1}^{\infty} x_{n}=X \text { and } \sum_{i=1}^{\infty} y_{n}=Y $$ If $$ \sum_{n=1}^{\infty} z_{n} $$ converges, then $$\lim _{n \rightarrow \infty} z_{n}=0 $$ Proof. Let $$S=\sum_{n=1}^{\infty} z_{n} $$ and $$S_{N}=\sum_{n=1}^{N} z_{n} . $$Then $$ \lim _{N \rightarrow \infty} z_{N}=\lim _{N \rightarrow \infty}\left(S_{N}-S_{N-1}\right)=S-S=0 $$ Let $\sum_{n=1}^{\infty} z_{n} $ be a series with $ \operatorname{sum} S $ and $ N^{th} $ $ \operatorname{sum} S_{N}= $ $ \sum_{n=1}^{N} z_{n} . $ The remainder $\rho_{N} $ after $ N terms Is ( \rho_{N}=S-S_{N}

Notice that series $\sum_{n=1}^{\infty} z_{n} $ converges to ( S ) if and only if sequence Want to know that. Is this proof right or can i shorten this proof? If anyother easy way to proof please provide

Please show your efforts on the question to avoid it being closed or heavily downvoted. See here to know how to ask a good question on this website. (Please take note that this website does not do your homework for you) — sai-kartik, May 17 '20 at 16:13
Does this answer your question? If a series converges, then the sequence of terms converges to $0$. – read the answers, they verbatim work with complex numbers as well. — Martin R, May 17 '20 at 16:19

score 1 · Answer 1 · answered May 17 '20 at 16:16

If $\sum_{n=1}^\infty z_n$ converges then the sequence $a_m = \sum_{n=1}^m z_n$ is convergent. So in particular, it is a Cauchy sequence, which means that for all $\epsilon > 0$ there exists some $M$ such that for all $m_1 \geq m_2 \geq M$:

$$|a_{m_1} - a_{m_2}| < \epsilon.$$

In particular this holds for $m_1 = m+1$ and $m_2 = m$ whenever $m \geq M$. Then

$$|a_{m_1} - a_{m_2}| = |\sum_{n=1}^{m+1} z_n -\sum_{n=1}^m z_n| = |z_{m+1}| < \epsilon.$$

So for all $m \geq M+1$ we have $|z_m| < \epsilon$. Since $\epsilon$ was arbitrary we have $\lim_{n \rightarrow \infty} z_n = 0$.

score 0 · Answer 2 · answered May 17 '20 at 17:38

Suppose $\sum\limits_{k=1}^{\infty} z_k$ converges. That means $\{\sum\limits_{k=1}^{n} z_k\}$ is a cauchy sequence. So for any $\epsilon >0$ there is an $N$ so that if $n,m > N$ then $|\sum\limits_{k=1}^{n} z_k - \sum\limits_{k=1}^{m} z_k|<\epsilon$. In particular for any $n > N+1$, $|\sum\limits_{k=1}^{n} z_k - \sum\limits_{k=1}^{n-1} z_k|=|z_n| < \epsilon$.

But if it were not true that $\lim\limits_{n\to \infty} z_n = 0$ then there would be a value $\epsilon_{death} > 0$ so that for any $N$ there will be an $n > N$ where $|z_n| > \epsilon_{death}$. That is contradictory to the result of paragraph 1. So it is not possilbe for $\sum\limits_{n=1}^{\infty} z_n$ and not have $\lim\limits_{n\to \infty} z_n = 0$.

I need toSolve this proof for complex numbers

2 Answers2