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\noindent {\bf Question}

\noindent Prove that if $x_n\rightarrow x$ as
$n\rightarrow\infty$, then $\frac{x_1 + \cdots + x_n}{n}
\rightarrow x$ as $n\rightarrow\infty$.

\medskip

\noindent {\bf Answer}

\noindent Since $\lim_{n\rightarrow\infty} x_n =x$, we have that
for each $\varepsilon
>0$, there exists $M$ so that $|x_n -x| <\frac{1}{3} \varepsilon$ for
$n >M$.  For any $m >0$ and $n >M$, we now have that
\begin{eqnarray*}
|x_{n+1}+\cdots +x_{n+m} -mx| & = & |x_{n+1} -x +\cdots  +x_{n+m}
-x| \\
 & \le & |x_{n+1}  -x| +\cdots +|x_{n+m} -x| \\
 & \le & m\frac{1}{3}\varepsilon.
\end{eqnarray*}
Dividing by $n+m$, we obtain that
\[ \left| \frac{1}{n+m}(x_{n+1}+\cdots +x_{n+m}) -\frac{m}{n+m} x
\right| \le \frac{m}{n+m} \frac{1}{3}\varepsilon
<\frac{1}{3}\varepsilon \] (since $\frac{m}{n+m} <1$).  Viewing
$n$ as fixed for the moment, choose $m$ so that both
$|\frac{m}{n+m}x -x| <\frac{1}{3}\varepsilon$ (which we can do
since $\lim_{m\rightarrow\infty} \frac{m}{n+m} =1$ for $n$ fixed)
and $\frac{1}{n+m} \left| x_1 +x_2+\cdots +x_n \right|
<\frac{1}{3}\varepsilon$ (which we can do since $x_1 +x_2+\cdots
+x_n$ is a constant when $n$ is fixed). Then,
\begin{eqnarray*}
\lefteqn{ \left| \frac{1}{n+m}(x_1+\cdots +x_{n+m}) -x \right|} \\
 & = & \left| \frac{1}{n+m}(x_1+\cdots +x_n) + \frac{1}{n+m}( x_{n+1}
 +\cdots +x_{n+m}) -\frac{m}{n+m} x +\frac{m}{n+m} x  -x \right| \\
 & \le & \left| \frac{1}{n+m}(x_1+\cdots +x_n) \right| + \left|
 \frac{1}{n+m}( x_{n+1}  +\cdots  +x_{n+m}) -\frac{m}{n+m} x \right| +
 \left| \frac{m}{n+m} x  -x \right| \\  & \le & \frac{1}{3}\varepsilon
 +\frac{1}{3}\varepsilon + \frac{1}{3}\varepsilon =\varepsilon
\end{eqnarray*}
for all $m >0$.  Since this is true for all $n >M$ and all $m >0$,
we have that $\left| \frac{1}{p}(x_1+\cdots +x_{p}) -x \right|
<\varepsilon$ for all $p > M$, as desired.

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