Let $\left\{ a_{n} \right\}$ be a sequence of real number 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 number such that $a_{1} = 2,a_{n + 1} = a_{n}^{2} - a_{n} + 1\forall n \in N$ and $S = \frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + \ldots + \frac{1}{a_{2019}}$, then

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Sol. [Key Concept: Iterate the recurrence to get the closed form.]$$a_{n + 1} - 1 = a_{n}\left( a_{n} - 1 ight)$$$$\begin{matrix} & \frac{1}{a_{n + 1} - 1} = \frac{1}{a_{n} - 1} - \frac{1}{a_{n}} \Rightarrow \frac{1}{a_{n}} = \frac{1}{a_{n} - 1} - \frac{1}{a_{n + 1} - 1} \ & \ \Rightarrow \ S = \frac{1}{a_{1}} + \frac{1}{a_{2}} + \ldots\frac{1}{a_{2019}} = \frac{1}{a_{1} - 1} - \frac{1}{a_{2020} - 1} = 1 - \frac{1}{a_{2020} - 1} \end{matrix}$$$$\begin{matrix} \Rightarrow \ a_{i} & \ > 1 \ ext{ }S & \ < 1 \ a_{2020} - 1 & \ = a_{2019}\left( a_{2019} - 1 ight) = a_{2019}a_{2018}\left( a_{2018} - 1 ight) \ & \ = a_{2019}a_{2018}a_{2017}\ldots a_{2}a_{1}\left( a_{1} - 1 ight) \ & \ = a_{1}a_{2}a_{3}\ldots a_{2019} \ (2019)^{2019} & \ < \left( \frac{\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + \ldots + \frac{1}{a_{2019}}}{a_{2020} - 1} ight)^{2019} < a_{1}a_{2}a_{3}\ldots a_{2019} = a_{2020} - 1 \ \Rightarrow \ \frac{1}{(2019)^{2019}} & ext{ }S > 1 - \frac{1}{(2019)^{2019}} \end{matrix}$$✔ (a) $S = 1 - \frac{1}{a_{2020} - 1} < 1$✘ (b) $S > 1 - \frac{1}{{(2019)}^{2019}}$✔ (c) $S > 1 - \frac{1}{{(2018)}^{2018}}$✔ (d) $S > 1 - \frac{1}{{(2019)}^{2019}}$

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