Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 129380 by mohammad17 last updated on 15/Jan/21

	Answered by physicstutes last updated on 15/Jan/21
	${a_n } = n^2 a_(n+1) = (n+1)^2 a_(n+1) −a_n = (n+1)^2 −n^2 = n^2 +2n+1−n^2 = 2n +1 n ∈ N ⇒ 2n + 1 > 0 hence a_(n+1) −a_n >0 ⇒ a_(n+1) > a_n thus {a_n } is an increasing. lim_(n→∞) a_n = lim_(n→∞) n^2 = + ∞ ← divergent hence a_n is not bounded$
	${a_{n}} = n^{2}$ $a_{n + 1} = {(n + 1)}^{2}$ $a_{n + 1} - a_{n} = {(n + 1)}^{2} - n^{2} = n^{2} + 2 n + 1 - n^{2} = 2 n + 1$ $n \in N \Rightarrow 2 n + 1 > 0 hence a_{n + 1} - a_{n} > 0 \Rightarrow a_{n + 1} > a_{n} thus$ ${a_{n}} is an increasing .$ $\lim_{n \to \infty} a_{n} = \lim_{n \to \infty} n^{2} = + \infty \leftarrow divergent$ $hence a_{n} is not bounded$
	Commented by mohammad17 last updated on 15/Jan/21

	$v e r y t h a n k s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum