a_(n+2) = (√(a_n ×a_(n+1) )) ∀ n≥1 , n ∈ N and here a_(1 ) = α and a_2 = β then prove that lim_(n→∞) a


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 198152 by universe last updated on 11/Oct/23

$a_{n + 2} = \sqrt{a_{n} \times a_{n + 1}} \forall n ⩾ 1, n \in N$ $and here a_{1} = α a n d a_{2} = β then$ $prove that \lim_{n \to \infty} a_{n + 2} = {(α \times β^{2})}^{1 / 3}$
	Answered by mr W last updated on 12/Oct/23

	$l e t b_{n} = \ln a_{n}$ $a_{n + 2} = \sqrt{a_{n + 1} \times a_{n}}$ $\ln a_{n + 2} = \frac{\ln a_{n + 1} + \ln a_{n}}{2}$ $2 \ln a_{n + 2} - \ln a_{n + 1} - \ln a_{n} = 0$ $\Rightarrow 2 b_{n + 2} - b_{n + 1} - b_{n} = 0 (r e c u r r e n c e r e l a t i o n)$ $2 p^{2} - p - 1 = 0 (c h a r a c t e r i s t i c e q u a t i o n)$ $(2 p + 1) (p - 1) = 0$ $\Rightarrow p = 1, - \frac{1}{2}$ $\Rightarrow b_{n} = A + B {(- \frac{1}{2})}^{n}$ $b_{1} = A + B (- \frac{1}{2}) = \ln a_{1} = \ln α . . . (i)$ $b_{2} = A + B {(- \frac{1}{2})}^{2} = \ln a_{2} = \ln β . . . (i i)$ $(i i) - (i) :$ $\frac{3 B}{4} = \ln β - \ln α$ $\Rightarrow B = \frac{4}{3} (\ln \frac{β}{α}) = \ln {(\frac{β}{α})}^{\frac{4}{3}}$ $\Rightarrow A = \frac{B}{2} + \ln α = \frac{2}{3} (\ln \frac{β}{α}) + \ln α = \ln {(α β^{2})}^{\frac{1}{3}}$ $\Rightarrow b_{n} = \ln {(α β^{2})}^{\frac{1}{3}} + {(- \frac{1}{2})}^{n} \ln {(\frac{β}{α})}^{\frac{4}{3}}$ $\Rightarrow b_{n} = \ln [{(α β^{2})}^{\frac{1}{3}} {(\frac{β}{α})}^{\frac{4}{3} {(- \frac{1}{2})}^{n}}] = \ln a_{n}$ $\Rightarrow a_{n} = {(α β^{2})}^{\frac{1}{3}} {(\frac{β}{α})}^{\frac{4}{3} {(- \frac{1}{2})}^{n}}$ $\Rightarrow \lim_{n \to \infty} a_{n} = {(α β^{2})}^{\frac{1}{3}} ✓$
	Commented by universe last updated on 12/Oct/23

	$t h a n k s s i r$
	Answered by Frix last updated on 12/Oct/23

	$a_{n + 2} = \sqrt{a_{n + 1} a_{n}}$ $a_{n} = e^{C_{1} {(- \frac{1}{2})}^{n} + C_{2}}$ $a_{1} = α \Leftrightarrow e^{- \frac{C_{1}}{2} + C_{2}} = α \Leftrightarrow - \frac{C_{1}}{2} + C_{2} = \ln α$ $a_{2} = β \Leftrightarrow e^{\frac{C_{1}}{4} + C_{2}} = β \Leftrightarrow \frac{C_{1}}{4} + C_{2} = \ln β$ $\Rightarrow$ $C_{1} = \frac{4}{3} \ln \frac{β}{α} \land C_{2} = \frac{1}{3} \ln α β^{2}$ $\Rightarrow$ $a_{n} = α^{\frac{1}{3} (1 - {(- 2)}^{2 - n})} β^{\frac{2}{3} (1 + {(- 2)}^{1 - n})}$ $\lim_{n \to \infty} a_{n} = α^{\frac{1}{3} (1 \pm 0)} β^{\frac{2}{3} (1 \mp 0)} = {(α β^{2})}^{\frac{1}{3}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum