<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.
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