Sequence Problem

Calculus Level 2

Let a recurrent sequence be defined as:

u 0 : = 2 u n + 1 : = 1 2 u n + 1 2 , n ∈ N \begin{aligned} u_0 & := 2 \\ u_{n+1} & := \dfrac 12 u_n + \dfrac 12, & n \in \mathbb N \end{aligned}

Find lim ⁡ n → ∞ u n \displaystyle \lim_{n \to \infty} u_n .


The answer is 1.

Indre Dan by Indre Dan , 24, Romania

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1 solution

Chew-Seong Cheong
Aug 18, 2016

Relevant wiki: Linear Recurrence Relations - Calculating Initial Terms

u n + 1 = 1 2 u n + 1 2 u n + 1 − 1 = 1 2 u n − 1 2 − 1 2 + 1 2 ( u n + 1 − 1 ) = 1 2 ( u n − 1 ) Let t n = u n − 1 t n + 1 = 1 2 t n Characteristic equation r = 1 2 ⟹ t n = a 2 n where a is a constant u n − 1 = a 2 n u n = a 2 n + 1 u 0 = 2 ⟹ a 2 0 + 1 = 2 ⟹ a = 1 ⟹ u n = 1 2 n + 1 ⟹ lim ⁡ n → ∞ u n = 0 + 1 = 1 \begin{aligned} u_{n+1} & = \frac 12 u_n + \frac 12 \\ u_{n+1} \color{#3D99F6}{-1} & = \frac 12 u_n \color{#3D99F6}{-\frac 12 - \frac 12} + \frac 12 \\ \left( \color{#3D99F6}{u_{n+1} -1} \right) & = \frac 12 \left(\color{#3D99F6}{u_{n} -1} \right) & \small \color{#3D99F6}{\text{Let }t_n = u_n -1} \\ \color{#3D99F6}{t_{n+1}} & = \frac 12 \color{#3D99F6}{t_{n}} & \small \color{#3D99F6}{\text{Characteristic equation }r=\frac 12} \\ \implies t_n & = \frac {\color{#3D99F6}{a}}{2^n} & \small \color{#3D99F6}{\text{where } a \text{ is a constant}} \\ u_n - 1& = \frac a{2^n} \\ u_n & = \frac a{2^n} + 1 \\ u_0 & = 2 \\ \implies \frac a{2^0} + 1 & = 2 \\ \implies a & = 1 \\ \implies u_n & = \frac 1{2^n} + 1 \\ \implies \lim_{n \to \infty} u_n & = 0 + 1 = \boxed{1} \end{aligned}

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