Bounding a Quadratic Recursion

Problem

A sequence {an}\{a_n\} has an>1a_n > 1, a1=2a_1 = 2, and

an+12an+1an2+1=0.a_{n+1}^2 - a_{n+1} - a_n^2 + 1 = 0.

1. Prove that n+74an<an+1n+2\dfrac{n+7}{4} \leqslant a_n < a_{n+1} \leqslant n + 2. 2. Prove that k=1n12ak23<1\displaystyle\sum_{k=1}^n \frac{1}{2a_k^2 - 3} < 1.

Answer

Solution

Difficulty8/10
TopicsRecursion, sequences, Estimation, Telescoping, inequality

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