Why does $\lim_{n\to\infty} z_n=A$ imply $\lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A$? 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.
 A: 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.
A: 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 
$$\lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A.$$
