Let $\left\{ a_{n} \right\}$ be a sequence of real numbers such that $a_{1} = 2,a_{n + 1} = a_{n}^{2

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Let $\left\{ a_{n} \right\}$ be a sequence of real numbers such that $a_{1} = 2,a_{n + 1} = a_{n}^{2} - a_{n} + 1$ for $n = 1$, $2,3,\ldots$ Let $S = \frac{1}{a_{1}} + \frac{1}{a_{2}} + \ldots + \frac{1}{a_{2018}}$, then

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Sol. [Key Concept: Telescoping / recurrence sum.]$$\frac{1}{a_{n + 1} - 1} = \frac{1}{a_{n}\left( a_{n} - 1 ight)} = \frac{1}{a_{n} - 1} - \frac{1}{a_{n}}$$$$\begin{matrix} \Rightarrow & & \frac{1}{a_{n} - 1} - \frac{1}{a_{n + 1} - 1} = \frac{1}{a_{n}} \ \Rightarrow \ & & \sum_{r = 1}^{2018}\mspace{2mu}\mspace{2mu}\frac{1}{a_{r}} = \frac{1}{a_{1} - 1} - \frac{1}{a_{2019} - 1} = 1 - \frac{1}{a_{2019} - 1} < 1 \ & & a_{2019} - 1 = a_{2018}\left( a_{2018} - 1 ight) = a_{2018}a_{2017}\left( a_{2017} - 1 ight) \ \Rightarrow \ & & a_{2019} - 1 = a_{2018}a_{2017}a_{2016}\ldots a_{2}a_{1}\left( a_{1} - 1 ight) \ & & = a_{1}a_{2}a_{3}\ldots a_{2018} \ \Rightarrow \ & & 2018 < \frac{1}{\frac{1}{a_{1}} + \frac{1}{a_{2}} + \ldots + \frac{1}{a_{2018}}} < \left( a_{1}a_{2}\ldots a_{2018} ight)\frac{1}{2018} \ \Rightarrow \ & & 1 - \frac{1}{a_{2019} - 1} > 1 - \frac{1}{(2018)^{2018}} \end{matrix}$$✘ (a) $S > 1 - \frac{1}{{(2018)}^{2018}}$✔ (b) $S > 1 - \frac{1}{{(2018)}^{2018}}$✔ (c) $S < 1$✘ (d) wrong bound

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