show that the sequence {a_n } defined recurssively by a_1 = (3/2) a_(n ) = (√(3a_(n−1 ) −2 )) for n≥2 converges and find its limit.


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 213468 by universe last updated on 06/Nov/24
$show that the sequence {a_n } defined recurssively by a_1 = (3/2) a_(n ) = (√(3a_(n−1 ) −2 )) for n≥2 converges and find its limit.$
$show that the sequence {a_{n}} defined$ $recurssively by a_{1} = \frac{3}{2}$ $a_{n} = \sqrt{{3 a}_{n - 1} - 2} for n ⩾ 2 converges$ $and find its limit .$
	Answered by issac last updated on 06/Nov/24
	$a_2 =(√(3a_1 −2))=(√((9/2)−2))=((√(10))/2) a_1 <a_2 ....true and let′s assume a_(k+1) <a_(k+2) , k∈Z (√(3a_k −2))<(√(3a_(k+1) −2)) ∴a_k <a_(k+1) ....true thus we can conclude that a_(n ) is ′′Monotonic Increase Sequence′′.. and Next we must show that sequence a_(n ) is Bounded or Not step 1. Let′s show that a_n inductively a_n ≤2 a_1 =(3/2)≤2 and if a_n ≤2 → a_(n+1) ≤2 a_(n+1) =(√(3a_n −2))≤2 ∴ a_n ≤2 so Sequence {a_n } is bounded lim_(n→∞) a_n =2 Q.E.D$
	$a_{2} = \sqrt{3 a_{1} - 2} = \sqrt{\frac{9}{2} - 2} = \frac{\sqrt{10}}{2}$ $a_{1} < a_{2} . . . . true$ $and {let}^{'} s assume a_{k + 1} < a_{k + 2}, k \in Z$ $\sqrt{3 a_{k} - 2} < \sqrt{3 a_{k + 1} - 2}$ $∴ a_{k} < a_{k + 1} . . . . true$ $thus we can conclude that a_{n} is$ $^{″} Monotonic Increase {Sequence}^{″} . .$ $and Next we must show that sequence$ $a_{n} is Bounded or Not$ $step 1 .$ ${Let}^{'} s show that a_{n} inductively a_{n} \leq 2$ $a_{1} = \frac{3}{2} \leq 2 and if a_{n} \leq 2 \to a_{n + 1} \leq 2$ $a_{n + 1} = \sqrt{3 a_{n} - 2} \leq 2$ $∴ a_{n} \leq 2 so Sequence {a_{n}} is bounded$ $\lim_{n \to \infty} a_{n} = 2 Q . E . D$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum