A Telescoping Reciprocal Sum

Problem

A sequence {an}\{a_n\} satisfies a1=43a_1 = \dfrac{4}{3} and

an+11=an2an,nN.a_{n+1} - 1 = a_n^2 - a_n, \quad n \in \mathbb{N}^*.

Let SnS_n be the sum of the first nn terms of {1an}\left\{\dfrac{1}{a_n}\right\}. Determine the pair of consecutive integers between which S2021S_{2021} lies, with proof.

Answer

Solution

Difficulty6/10
TopicsRecursion, sequences, Telescoping, inequality

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