Soient a et b deux re^� els. Pour tout n ∈ N on pose u_n =(√n)+a(√(n+1))+b(√(n+2)). 1. Ve^� rifier que la suite (u_n ) tend vers 0 si et seulement si a+b=−1. 2. De^� terminer a et b pour que la se^� rie Σu


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 173795 by Ar Brandon last updated on 18/Jul/22

$Soient a et b deux r \overset{´}{e e l s} . Pour tout n \in N on pose$ $u_{n} = \sqrt{n} + a \sqrt{n + 1} + b \sqrt{n + 2} .$ $1 . V \overset{´}{e r i f i e r} que la suite (u_{n}) tend vers 0 si et seulement si a + b = - 1 .$ $2 . D \overset{´}{e t e r m i n e r} a et b pour que la s \overset{´}{e r i e} Σ u_{n} soit convergente .$
Commented by Ar Brandon last updated on 18/Jul/22

$Q 2 .$
	Answered by mindispower last updated on 18/Jul/22

	$1 U_{n} = \sqrt{n} + a \sqrt{n} (\sqrt{1 + \frac{1}{n}}) + b \sqrt{n} (\sqrt{1 + \frac{2}{n}})$ $= \sqrt{n} (1 + a \sqrt{1 + \frac{1}{n}} + b \sqrt{1 + \frac{2}{n}})$ $\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{1}{8} x^{2} + o (x^{2})$ $U_{n} = \sqrt{n} (1 + a + \frac{a}{2 n} - \frac{a}{8 n^{2}} + b + \frac{2 b}{2 n} - \frac{4 b}{8 n^{2}})$ $U_{n} \to 0 \Leftrightarrow 1 + a + b = 0 \Rightarrow a + b = - 1$ $2) U_{n} \to C v \Leftrightarrow$ $w h e n n \to \infty U_{n} \sim V_{n} a n d Σ V_{n} < \infty$ $U_{n} = \sqrt{n} (1 + a + b + \frac{a + 2 b}{2 n} - \frac{a + 4 b}{8 n^{2}} + o (\frac{1}{n^{2}}))$ $\sqrt{n} (1 + a + b + \frac{a + 2 b}{2 n} - \frac{a + 4 b}{8 n^{2}} + o (\frac{1}{n^{2}})) c v$ $i f a n d o n l y i f$ $1 + a + b = 0$ $a + 2 b = 0 \Rightarrow b = 1, a = - 2$ $U_{n} = \sqrt{n} - 2 \sqrt{1 + n} + \sqrt{n + 2}$
	Commented by Ar Brandon last updated on 18/Jul/22
	Merci
	Commented by Tawa11 last updated on 19/Jul/22

	$Great sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum