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Problem 6: (1) The inequality. (2)Show that $(\sigma_ n)\to S$. (3) Evaluate a limit. 

[Solution] (1) Let $\varepsilon\textgreater 0$. By definition there is an $N_\varepsilon$ such that for all $n\textgreater N_\varepsilon$,
\begin{align}
s_1 = &s_1 = s_1\notag\\
&\vdots\notag\\
s_{N_\varepsilon} = &s_{N_\varepsilon} = s_{N_\varepsilon}\notag\\
S-\varepsilon \textless &s_{N_\varepsilon +1} \textless S+\varepsilon\notag\\
&\vdots\notag\\
S-\varepsilon \textless &s_n \textless S+\varepsilon\notag\\
\end{align}
Summing up, we get 
\begin{align}
\frac{s_1 + ..+ s_{N_{\varepsilon } } }{n} + \frac{ (n- N_{\varepsilon })(S-\varepsilon )}{n} \textless \sigma _n \textless  \frac{s_1 + ..+ s_{N_{\varepsilon  } } }{n} + \frac{ (n- N_{\varepsilon })(S+\varepsilon )}{n}. \notag
\end{align}

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$\text{ }$

(2)
\begin{proof}

[1] Let $0\textless\epsilon ' \textless 1$. Then by (1), we find an $N_{\epsilon ' } $ such that $\forall n\geq N_{\epsilon '}$,
\begin{align}
\frac{s_1 + ..+ s_{N_{\epsilon ' } } }{n} + \frac{ (n- N_{\epsilon ' })(S-\epsilon ' )}{n} \textless \sigma _n \textless  \frac{s_1 + ..+ s_{N_{\epsilon ' } } }{n} + \frac{ (n- N_{\epsilon ' })(S+\epsilon ' )}{n}. \notag
\end{align}
Since $\lim_{n\to\infty} \frac{s_1 + ..+ s_{N_{\epsilon ' } } }{n} =0$ and $\lim_{n\to\infty}  \frac{ n- N_{\epsilon ' }}{n} =1$, there is an $N ' _{\epsilon ' }\textgreater N_{\epsilon ' }$ such that for all $n\geq N'_{\epsilon ' } $, 
\begin{align}
-\epsilon ' \textless \frac{s_1+...+s_{N_{\epsilon '}}}{n} &\textless \epsilon ' \notag\\
1-\epsilon ' \textless \frac{n-N_{\epsilon ' } }{n} &\textless 1+\epsilon ' \notag
\end{align}
At this time, 
\begin{align}
\sigma _n &\textless \epsilon ' + (1\pm\epsilon ')(S+\epsilon ' )\notag\\
&=\epsilon ' + S +  (1\pm S)\epsilon ' \pm \epsilon ' \ ^2  \notag\\
&\leq \epsilon ' + S + |1\pm S|\epsilon ' + \epsilon ' \notag\\
&\leq \epsilon ' + S + (1+|S|) \epsilon ' + \epsilon ' \notag\\
&= S +(3+ |S|)\epsilon '\notag
\end{align}
\begin{align}
\sigma _n &\textgreater  -\epsilon ' + (1\mp\epsilon ')(S-\epsilon ' )\notag\\
  &= -\epsilon '+S-  (1\pm S)\epsilon '  \pm \epsilon ' \  ^2  \notag\\
&\geq -\epsilon ' + S - |1\pm S|\epsilon ' - \epsilon ' \notag\\
&\geq -\epsilon ' + S - (1+|S|) \epsilon ' - \epsilon ' \notag\\
&= S -(3+ |S|)\epsilon '\notag
\end{align}
where the choice of $\pm$ and $\mp$ depends on the sign of $S+\epsilon'$ and $S-\epsilon ' $. Hence 
\begin{align}
|\sigma_n - S| \textless (3+|S|)\epsilon ' \notag
\end{align}

[2] Next, we prove $\lim_{n\to\infty} \sigma _n = S$. Let $\varepsilon \textgreater 0 $. Choose $\epsilon ' = min \{\frac{\varepsilon}{3+|S|},\frac{1}{2}\}$. Then there is an $N_{\epsilon ' }$ and an $N' _ {\epsilon ' }\textgreater N_{\epsilon ' }$, such that for any $n\geq N' _ {\epsilon ' }$, 
\begin{align}
|\sigma _n - S| &\textless (3+|S|) \epsilon ' \notag\\
& \leq (3+ |S|) \frac{\varepsilon}{3+|S|} = \varepsilon\notag
\end{align}
\end{proof}

[Note] (a) The limit $\lim_{n\to\infty}\sigma_n$ is call $\mathit{Cesaro}$ $\mathit{Sum}$. (b) If we have the technique of upper and lower limits, the proof can be simplified. 


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(3) Let $\tau_n = 1+\frac{1}{2} + ...+ \frac{1}{2^{n-1}}$, $\varsigma ' _n = \frac{1}{n} \{0\cdot \frac{1}{2^0} + 1\cdot\frac{1}{2^1} + ... + (n-1)\frac{1}{2^{n-1}} \}$, $s_n = \frac{n-1}{2^{n-1}}$. Then 
\begin{align}
\sigma_n = \tau _n - \varsigma ' _n\notag
\end{align}
Since $s_n \to 0$ (check !), we get  $\varsigma ' _n \to 0$. Since we also have $\tau_n\to 2$, then $\sigma_n \to 2 $ as well.

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