Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm self-studying a bit of complex analysis, and I'm attempting to figure out the following.

Suppose $\lim_{n\to\infty}z_n=A$. How can I show that $$ \lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A. $$

Is there a clever way to write the limit to make it more approachable? Thank you.

share|cite|improve this question
It'll also be helpful, – Ehsan M. Kermani Jan 30 '12 at 4:11
Thanks for the link, @ehsanmo. – Dedede Jan 30 '12 at 4:19
up vote 6 down vote accepted

Let $\epsilon>0$. Since $\lim_{n\to\infty}z_n=A$, there exists an positive integer $N$ such that $$\tag{1}|z_n-A|<\frac{\epsilon}{2}\mbox{ whenever }n\geq N.$$ For this fixed $N$, we can find another positive integer $N'$ such that $$\tag{2}\sum_{k=1}^{N-1}|z_i-A|\leq \frac{N'\epsilon}{2}.$$

Hence, if $m\geq\max\{N, N'\}$, then $$\left|\frac{1}{m}(z_1+\cdots+z_m)-A\right|=\left|\frac{1}{m}\sum_{k=1}^{m}(z_i-A)\right|$$ $$\leq\left|\frac{1}{m}\sum_{k=1}^{N-1}(z_i-A)\right|+\left|\frac{1}{m}\sum_{k=N}^{m}(z_i-A)\right|$$ $$\leq\frac{1}{m}\sum_{k=1}^{N-1}|z_i-A|+\frac{1}{m}\sum_{k=N}^{m}|z_i-A|$$ $$\leq \frac{\epsilon}{2}+\frac{1}{m}(m-N+1)\frac{\epsilon}{2}\leq \epsilon$$ where we have used $(1)$ and $(2)$ in the last inequality. This proves that $$\lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A.$$

share|cite|improve this answer
Thank you Paul, I find this answer very clear and easy to understand. – Dedede Jan 30 '12 at 2:50
@Dedede: You are welcome. – Paul Jan 30 '12 at 3:45

We have that


so it suffices to prove the statement when $A=0$. The $\epsilon$-$N$ definition of the limit tells us that for any positive $\epsilon>0$, there is an $N$ such that $n>N\implies |z_n|<\epsilon$, which subsequently implies


$$\le \left|\frac{z_1+\cdots+z_N}{k+N}\right|\,+\,\left|\frac{z_{N+1}+\cdots+z_{N+k}}{k+N}\right|\le \left|\frac{z_1+\cdots+z_N}{k+N}\right|+\epsilon\stackrel{k\to\infty}{\longrightarrow}\epsilon.$$

(We also used the triangle inequality above.) Now take $\epsilon\to0^+$ and use the squeeze theorem.

share|cite|improve this answer
Thanks again for your help, anon. – Dedede Jan 30 '12 at 2:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.