A Two-Sided Series Bound

Problem

A sequence {an}\{a_n\} satisfies a1=1a_1 = 1 and

nan+1=(n+2)an+n,na_{n+1} = (n+2)a_n + n,

and bn=ann(n+1)b_n = \dfrac{a_n}{n(n+1)}.

1. Find bnb_n and ana_n. 2. Prove that

n21a2+1a3++1an+1+b12+b22++bn22n24n+4n12n.\frac{n}{2} \leqslant \frac{1}{a_2} + \frac{1}{a_3} + \cdots + \frac{1}{a_{n+1}} + b_1^2 + b_2^2 + \cdots + b_n^2 \leqslant \frac{2n^2 - 4n + 4\sqrt{n} - 1}{2n}.

Answer

Solution

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

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