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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.

Dedede
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  • It'll also be helpful, http://www.math.columbia.edu/~nironi/stolz-cesaro.pdf – Ehsan M. Kermani Jan 30 '12 at 04:11
  • Thanks for the link, @ehsanmo. – Dedede Jan 30 '12 at 04:19
  • There are also several posts about the same problem for real sequences. Probably not all of them, but many of the proofs which work for real sequences also work for complex sequences. http://math.stackexchange.com/questions/207910/prove-convergence-of-the-sequence-z-1z-2-cdots-z-n-n-of-cesaro-means http://math.stackexchange.com/questions/210681/if-a-n-to-ell-then-hat-a-n-to-ell http://math.stackexchange.com/questions/248116/arithmetic-mean-of-a-sequence-converges http://math.stackexchange.com/questions/930436/a-result-on-sequences-x-n-to-x-implies-fracx-1-dotsx-nn-to-x-without-u – Martin Sleziak Sep 22 '16 at 03:47

2 Answers2

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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 $$\lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A.$$

Paul
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We have that

$$\frac{z_1+\cdots+z_n}{n}-A=\frac{(z_1-A)+\cdots+(z_n-A)}{n},$$

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

$$\left|\frac{z_1+\cdots+z_{k+N}}{k+N}\right|=\left|\frac{z_1+\cdots+z_N}{k+N}\,+\,\frac{z_{N+1}+\cdots+z_{N+k}}{k+N}\right|$$

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

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

anon
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