How find this seqence of $(n-1)a_{n}<1$

Question

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

are $a_0$ and $a_1$ given to you? – Prahlad Vaidyanathan Oct 04 '13 at 11:42 — Prahlad Vaidyanathan, Oct 04 '13 at 11:42
$a_{0},a_{1}$ is or any postive numbers, – math110 Oct 04 '13 at 11:59 — math110, Oct 04 '13 at 11:59

score 3 · Accepted Answer · answered Oct 05 '13 at 02:23

$$1=\frac{a_{k-1}-a_{k+1}}{(a_{k-1}+a_{k})(a_{k}+a_{k+1})}=\frac{1}{a_{k}+a_{k+1}}-\frac{1}{a_{k-1}+a_{k}}$$

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.

1 Answers1