a_(n+1 ) = a_n + (√(a_n ^2 + 1)) , a_0 = 0 find a


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 199309 by universe last updated on 01/Nov/23

$a_{n + 1} = a_{n} + \sqrt{a_{n}^{2} + 1}, a_{0} = 0$ $find a_{n} = ? ?$
	Answered by TheHoneyCat last updated on 01/Nov/23

	$a_{n}^{2} > 0$ $\Rightarrow a_{n}^{2} + 1 > 1$ $\Rightarrow \sqrt{a_{n}^{2} + 1} > 1$ $\Rightarrow a_{n + 1} > 1$ $Thus (a_{n}) is strictly increasinng$ $it is therefore either convergent to l \in R$ $and l = l + \sqrt{l^{2} + 1}$ $or l = + \infty$ $the eqution on l gives l^{2} = - 1 wich is not$ $possible in R$ $thus : a_{n} \underset{n \to \infty}{\to} \infty_{◼}$
	Commented by TheHoneyCat last updated on 01/Nov/23
	woops��. I misread the question. sorry...
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum