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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

2 Answers 2

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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!

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    I seem to be puzzled. $\sum (z_k - z) = (z_0 -z) + (z_1-z)+...+(z_n-z)=z_0+z_1+...z_n - (n+1)z \ne \sum z_k -z = (z_0+z_1+...+z_n) -z$. – 2017-02-17
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    No need to be puzzled. Divide by $n+1$ and we obtain $\frac{1}{n+1}\sum_{m=0}^n(z-z_m)=\frac{z_0+z_1+z_2+\cdots +z_n}{n+1}-z$. And we want to show that this approaches $0$ as $n\to \infty$. ;-)) – 2017-02-17
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    Oops, I paid no attention to the existence of $\frac{1}{n+1}$... – 2017-02-17
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Hint:

$$\left|\frac{1}{n+1}\sum_{k=0}^n z_k - z \right| \leqslant \frac{1}{n+1} \sum_{k=0}^N |z_k - z|+ \frac{1}{n+1} \sum_{k=N+1}^n |z_k -z|$$

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    This is simple and awesome. Thanks! – 2017-02-17
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    @sequence: You're welcome – 2017-02-17
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    Can you please show how $\left|\sum_{k=0}^n z_k - z \right| \leqslant \sum_{k=0}^n |z_k - z|$? @RRL – 2017-02-17
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    Also, this actually doesn't seem to help even if it held... We can't make $|z_k-z|<\varepsilon$ for $k – 2017-02-17
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    You don't need to. With $N$ fixed the first sum on the RHS goes to zero as $n \to \ infty$. So for n sufficiently large the entire term is made smaller than $\epsilon/2$. – 2017-02-17
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    @sequence: $\frac{1}{n+1}\sum_{k=0}^n z_k - z = \frac{1}{n+1}\sum_{k=0}^n (z_k - z)$. – 2017-02-17