Series of terms of convergent sequence converges Let $z_n$ be a sequence, such that $\lim\limits_{n\to \infty}z_n = z$. Then $\lim\limits_{n\to \infty}(z_0+...+z_n)/(n+1)=z$.
Here's what I was trying to do:
We are given that $\forall \varepsilon>0, \exists N>0$ such that $|z_n-z|<\varepsilon$ whenever $n>N$.
Now, $\left|\sum\limits_{k=0}^n \frac{1}{n+1}z_k -z \right|\le \left| \sum\limits_{k=0}^n \frac{1}{n+1}\max\limits_{k\in\{0,...,n\} }\{z_k\} -z \right|\le \left|  \frac{1}{n+1}(n+1)\max\limits_{k\in\{0,...,n\} }\{z_k\} -z \right|=\left|  \max\limits_{k\in\{0,...,n\} }\{z_k\} -z \right|$
The problem here is that $\max\limits_{k\in\{0,...,n\} }\{z_k\}$ may be for $k<N$, and so this inequality will not be less than $\varepsilon$. What do I do?
 A: Hint:
$$\left|\frac{1}{n+1}\sum_{k=0}^n z_k  - z \right| \leqslant \frac{1}{n+1} \sum_{k=0}^N |z_k - z|+ \frac{1}{n+1} \sum_{k=N+1}^n |z_k -z|$$
A: Let $\epsilon>0$ be given.  Then, there exists a number $N(\epsilon)$ such that whenever $n>N(\epsilon)$, $|z-z_n|<\epsilon/2$.  Then, fixing $N(\epsilon)$ we can write
$$\begin{align}
\left|\frac{1}{n+1}\sum_{m=0}^{n}(z-z_m)\right|&=\left|\frac{1}{n+1}\sum_{m=0}^{N(\epsilon)}(z-z_m)+\frac{1}{n+1}\sum_{N(\epsilon)+1}^n (z-z_m)\right|\\\\
&\le \color{blue}{\frac{1}{n+1}\sum_{m=0}^{N(\epsilon)}|z-z_m|}+\color{red}{ \frac{1}{n+1}\sum_{m=N(\epsilon)+1}^{n}|z-z_m|}\\\\
&\le \color{blue}{\frac{N(\epsilon)+1}{n+1}\max_{0\le m\le N(\epsilon)}\left|z-z_m\right|}+\color{red}{\frac{\epsilon}{2}\left(1-\frac{N(\epsilon)-1}{n+1}\right)}\\\\
&<\color{blue}{\frac\epsilon2}+\color{red}{\frac\epsilon2}\\\\
&=\epsilon
\end{align}$$
whenever $\displaystyle n>\max\left(N(\epsilon),\frac{2(N(\epsilon)+1)}{\epsilon}\,\max_{0\le m\le N(\epsilon)}\left|z-z_m\right|\right)$.
And we are done!
