Consider the sequence $\{a_k\}_{k \geq 1}$ defined by $a_1 = 1$, $a_2 = \frac{1}{2}$ and \[ a_{k + 2} = a_k + \frac{1}{2}a_{k + 1} + \frac{1}{4a_ka_{k + 1}}\ \textrm{for}\ k \geq 1. \] Prove that \[ \frac{1}{a_1a_3} + \frac{1}{a_2a_4} + \frac{1}{a_3a_5} + \cdots + \frac{1}{a_{98}a_{100}} < 4. \]
Problem
Source: Baltic Way 2005/3
Tags: algebra proposed, algebra
e.lopes
11.11.2005 16:36
here $\frac{1}{4a_ka_{k+1}}$ are useless, no? see that $a_{k+2} > a_k + \frac{a_{k+1}}{2}$ $\Longrightarrow \frac{1}{a_ka_{k+2}} < \frac{2}{a_ka_{k+1}} - \frac{2}{a_{k+1}a_{k+2}}$. so, $S < \frac{2}{a_1a_2} - \frac{2}{a_2a_3} + \frac{2}{a_2a_3} - ... < \frac{2}{a_1a_2} = 4$.
Arne
11.11.2005 19:13
That's right. The problem is somewhat tricky, I don't think any team solved it.
sinajackson
23.05.2007 14:01
I know this solution is not a nice one,e.lopes's solution is really good and enough...but me and my friend solved it this way: we proved that for each $n\geq1$ we have $a_{n}\geq\frac{n-1}{2}$ then: $\sum^{98}_{k=1}\frac{1}{a_{k}a_{k+2}}=\frac{1}{a_{1}a_{3}}+\sum^{98}_{k=2}\frac{1}{a_{k}a_{k+2}}\leq\frac{4}{7}+\sum^{98}_{k=2}\frac{4}{(k-1)(k+1)}=\frac{4}{7}+2\sum^{98}_{k=2}\left(\frac{1}{k-1}-\frac{1}{k}+\frac{1}{k}-\frac{1}{k+1}\right) =\frac{4}{7}+2\left(1-\frac{1}{98}+\frac{1}{2}-\frac{1}{99}\right)<4$