(1)证明:对任意x1,x2∈R,有
|f(x1)−f(x2)|=|
1+
x21−
1+
x22|=|
x21−
x22
1+
x21+
1+
x22|=
|x1−x2|•|x1+x2|
1+
x21+
1+
x22.…2分
由|f(x1)-f(x2)|≤L|x1-x2|,即
|x1−x2|•|x1+x2|
1+
x21+
1+
x22≤L|x1-x2|.
当x1≠x2时,得L≥
|x1+x2|
1+
x21+
1+
x22.
∵
1+
x21>|x1|,
1+
x22>|x2|,且|x1|+|x2|≥|x1+x2|,
∴
|x1+x2|
1+
x21+
1+
x22<
|x1+x2|
|x1|+|x2|≤1.…4分
∴要使|f(x1)-f(x2)|≤L|x1-x2|对任意x1,x2∈R都成立,只要L≥1.
当x1=x2时,|f(x1)-f(x2)|≤L|x1-x2|恒成立.
∴L的取值范围是[1,+∞).…5分
(2)证明:①∵an+1=f(an),n=1,2,…,
故当n≥2时,|an-an+1|=|f(an-1)-f(an)|≤L|an-1-an|=L|f(an-2)-f(an-1)|≤L2|an-2-an-1|≤…≤Ln-1|a1-a2|
∴
n
k=1|ak−ak+1|=|a1−a2|+|a2−a3|+|a3−a4|+…+|an−an+1|≤(1+L+L2+…+Ln-1)|a1-a2