Exercise 2.3.11

Question

[Cesaro Means]
  \begin{enumerate}[label=(\alph*)] 
  \item Show that if $\left(x_{n}\right)$ is a convergent sequence, then the sequence given by the averages
    $$
    y_{n}=\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}
    $$
    also converges to the same limit.
  \item Give an example to show that it is possible for the sequence $\left(y_{n}\right)$ of averages to converge even if $\left(x_{n}\right)$ does not.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Let $D = \sup\{|x_n - x| : n \in \mathbf N\}$ and let $0<\epsilon N_1$ giving $$ |y_n - x| \le \left|\frac{|x_1-x| + \dots + |x_{N_1}-x| + \dots + |x_n-x|}{n} ight| \le \left|\frac{N_1 D + (n-N_1)\epsilon/2}{n} ight| $$ Let $N_2$ be large enough that for all $n > N_2$ (remember $0 < \epsilon < D$ so $(D-\epsilon/2)>0$.) $$ 0 < \frac{N_1(D - \epsilon/2)}{n} < \epsilon/2 $$ Therefor $$ |y_n - x| \le \left|\frac{N_1(D - \epsilon/2)}{n} + \epsilon/2 ight| < \epsilon $$ Letting $N = \max\{N_1, N_2\}$ completes the proof as $|y_n - x| < \epsilon$ for all $n > N$. (Note: I could have used any $\epsilon' < \epsilon$ instead of $\epsilon/2$, I just needed some room.) \item $x_n = (-1)^n$ diverges but $(y_n) o 0$. \end{enumerate}

Richard Ganaye · Answer

\begin{proof}
\begin{enumerate}
\item[a)] 
Write $x =\lim\limits_{n 	o \infty} x_n$.

Let $\varepsilon$ be any positive real number. There is some $N \in \mathbf{N}$ such that, for all $n \in \mathbf{N}$,
$$ n\geq N \Rightarrow |x_n - x | < \varepsilon/2.$$
Write $M = |x_1 - x| + \cdots + |x_{N}-x|$. For all $n\in \mathbf{N}$ such that $n\geq N$,
\begin{align*}
|y_n - x| &= \left| \frac{x_1+\cdots + x_n}{n} - x ight|\
&= \left | \frac{(x_1 -x) + \cdots + (x_n - x)}{n} ight |\
&\leq \frac{|x_1 - x| + \cdots + |x_{N}-x|}{n} + \frac{|x_{N+1} - x| + \cdots + |x_n -x|}{n} \
&\leq \frac{M}{n} + \frac{(n-N)}{n}\frac{\varepsilon}{2}\
&\leq \frac{M}{n} + \frac{\varepsilon}{2}\qquad (	ext{because } (n-N)/n \leq  1).
\end{align*}
Moreover, there exists $N_1 \geq N$ such that for all $n \geq N_1$,
$$\frac{M}{n} < \frac{\varepsilon}{2},$$
(take $N_1= \max(N, \left\lfloor \frac{2M}{\varepsilon} ightfloor + 1$).
This shows that
$$\forall \varepsilon>0,\  \exists N_1 \in \mathbf{N},\ \forall n  \in \mathbf{N},\  n\geq N_1\Rightarrow  |y_n - x | < \varepsilon,$$
so that $\lim\limits_{n 	o \infty} y_n = x.$
\item[b)] $x_n = (-1)^n$ gives such an example.
\end{enumerate}
\end{proof}

Exercise 2.3.11

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Exercise 2.3.11

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