※ 引述《secretnetman (秘密網路人)》之銘言：
: A(n) -> a as n趨於無窮大
: 證明:
: [A(1)+...+A(n)]/n -> a as n 趨於無窮大
: 典型的高微問題
: 但小弟實力太差 希望高手解惑
Assume e>0.
Then there exists integer n1 such that if n>n1 then |A(n)-a|<e/2.
Let R=A(1)+A(2)+...+A(n1)-n1*a.
Then there exists integer n2 such that |R/n2|<e/2.
Let N=max(n1,n2).
We have for n>N,
|[A(1)+...+A(n)]/n-a|
<= | ([A(1)+...+A(n1+1)]-n1*a)/n | + | [A(n1+1)+...+A(n)-(n-n1)*a]/n |
< e/2+e/2
=e.
Since e is arbitrarily chosen,
we get that [A(1)+...+A(n)]/n→a as n→infinity.
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