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Calculus1 8 Online
OpenStudy (anonymous):

Show that the sequence an defined by: a1= √ 2 & an=√ 2+an-1,n>1 converges

geerky42 (geerky42):

\(a_n = \sqrt{2+a_{n-1}}\), right? In that case, \(a_2 = \sqrt{2+\sqrt2},~a_3 = \sqrt{2+\sqrt{2+\sqrt2}},~\cdots\) So \(S = \displaystyle\lim_{n\to\infty}a_n \Longrightarrow S = \sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=\sqrt{2+S}\) So you have \(S = \sqrt{2+S}\). Solve for S, then you can see that as n approaches to infinity, S approaches to certain value, hence it converges. Hope that helps.

OpenStudy (jhannybean):

Its like inception

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