TITLE: How find this seqence of $(n-1)a_{n}<1$
QUESTION [5 upvotes]: let sequence $\{a_{n}\}$ are positive numbers,and such

$$(a_{k-1}+a_{k})(a_{k}+a_{k+1})=a_{k-1}-a_{k+1},\forall k\in N^{+},k\ge 2$$
show that: $(n-1)a_{n}<1,n\ge 2$

This problem is my frend ask me,and I think use introduction it
Maybe this problem have nice methods,Thank you

REPLY [3 votes]: $$1=\frac{a_{k-1}-a_{k+1}}{(a_{k-1}+a_{k})(a_{k}+a_{k+1})}=\frac{1}{a_{k}+a_{k+1}}-\frac{1}{a_{k-1}+a_{k}}$$
Let $b_k=\frac{1}{a_{k-1}+a_{k}}$.
Then you get $b_{k+1}-b_k=1 \Rightarrow b_{k+1}=1+b_k$.
It follows by induction that 
$$b_{k}=k-1+b_1$$
Or 
$$a_{k-1}+a_k=\frac{1}{k-1+b_1}$$
Then, as $b_1 >0$ we have
$$a_n< a_{n-1}+a_n=\frac{1}{n-1+b_1} < \frac{1}{n-1} $$
Which is exactly what you need.