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Question Number 196913 by ERLY last updated on 02/Sep/23

s o i t {_{r_{n + 1} = r_{n} / (2 + r_{n}^{2})}^{r_{0} = 1}

d e m o n t r e r s a n s r e c u r r e n c e q u e r_{n} > 0

d e m o n t r e r p a r r e c u r r e n c e q u e r_{n + 1} \leq \frac{1}{2} r_{n}

d e m o n t r e r s a n s r e c u r r e n c e q u e r_{n} \leq {(\frac{1}{2})}^{n}

∙ e r l y r o l v i n s t ∙

Answered by aleks041103 last updated on 03/Oct/23

1 .

b y i n d u c t i o n

r_{0} = 1 > 0

l e t r_{k} > 0, f o r s o m e k \in N

r_{k + 1} = \frac{r_{k}}{2 + r_{k}^{2}} > 0

\Rightarrow r_{n} > 0

2 .

2 + r_{n}^{2} ⩾ 2 \Rightarrow \frac{1}{2} ⩾ \frac{1}{2 + r_{n}^{2}}

r_{n} > 0 \Rightarrow \frac{r_{n}}{2} ⩾ \frac{r_{n}}{2 + r_{n}^{2}} = r_{n + 1}

\Rightarrow r_{n + 1} ⩽ \frac{1}{2} r_{n} \to g e o m e t r i c p r o g r e s s i o n

3 .

\Rightarrow r_{n} ⩽ {(\frac{1}{2})}^{n} r_{0} = {(\frac{1}{2})}^{n}

4 .

0 < r_{n} ⩽ {(\frac{1}{2})}^{n}

\Rightarrow \underset{n \to \infty}{l i m} 0 ⩽ \underset{n \to \infty}{l i m} r_{n} ⩽ \underset{n \to \infty}{l i m} {(\frac{1}{2})}^{n}

\Rightarrow \underset{n \to \infty}{l i m} r_{n} = 0