epsilon-delta proofs 5

<exercises>

1. When $n\rightarrow \infty $, what does $f_n=\frac{1}{n}$ ($n\in\mathbb{N}$)  approach?

For any $\epsilon$, we are always able to set $N>\frac{1}{\epsilon}$. That is, $\frac{1}{N}<\epsilon $. Therefore, if $n\geq N$,
\[ 0<\frac{1}{n}\leq\frac{1}{N}<\epsilon  \]
Hence, we proved
\[ \frac{1}{n}\rightarrow 0\quad (n\rightarrow \infty ) \]


2. Find $f_n=\frac{a_1+a_2+\cdots +a_n}{n}$, when $n\rightarrow \infty $ and $a_n\rightarrow a$.

(using the result of exercise-1)
\begin{eqnarray*}
f_n &=& \frac{(a_1-a)+(a_2-a)+\cdots +(a_n-a)+na}{n}\\
 &=& (a_1-a)\frac{1}{n}+(a_2-a)\frac{1}{n}+\cdots +(a_n-a)\frac{1}{n}+a\rightarrow a\quad \left(\frac{1}{n}\rightarrow 0\right)
\end{eqnarray*}

(epsilon-delta proof)
\begin{eqnarray*}
|f_n-a| &=& \left| \frac{(a_1-a)+(a_2-a)+\cdots +(a_n-a)}{n}\right| \\
 &=& \left| \frac{(a_1-a)+\cdots +(a_m-a)+(a_{m+1}-a)+\cdots +(a_n-a)}{n}\right|
\end{eqnarray*}
Based on [Def-1], for a $\epsilon>0$, we are able to choose a $N$ such that if $n\geq N$, $|a_n-a|<\frac{\epsilon}{2}$. Similarly, we can also choose $N_1>N$ such that if $n\geq N_1$,
\[  \left|  \frac{\sum_{i=1}^m (a_i-a)}{n} \right|<\frac{\epsilon}{2} \]
Of course, for this $n$, $|a_n-a|<\frac{\epsilon}{2}$ is clear.  Then
\[  \left|  \frac{\sum_{i=m+1}^n (a_i-a)}{n} \right|<\frac{(n-m)}{n}\frac{\epsilon}{2} < \frac{\epsilon}{2} \]
Hence
\begin{eqnarray*}
|f_n-a| &<&\frac{\epsilon}{2}+ \frac{\epsilon}{2}=\epsilon \\
\end{eqnarray*}
A proof is completed.