Let&nbsp;a1,b1&nbsp;be two real numbers such that&nbsp;0&lt;a1&lt;b1. For&nbsp;n≥1, we define&nbsp;an+1=(anbn)12&nbsp;and&nbsp;bn+1=(an+bn)2a) Prove that the sequence&nbsp;{an}n∈N&nbsp;is monotonically increasing and that the sequence&nbsp;{bn}n∈N&nbsp;is monotonically decreasing.b) Show that the sequences&nbsp;{an}&nbsp;and&nbsp;{bn}&nbsp;are bounded.c) Deduce that the two sequences converge and prove that they converge to the same limit.

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Let&amp;nbsp;a1,b1&amp;nbsp;be two real numbers such that&amp;nbsp;0&amp;lt;a1&amp;lt;b1. For&amp;nbsp;n≥1, we define&amp;nbsp;an+1=(anbn)12&amp;nbsp;and&amp;nbsp;bn+1=(an+bn)2a) Prove that the sequence&amp;nbsp;{an}n∈N&amp;nbsp;is monotonically increasing and that the sequence&amp;nbsp;{bn}n∈N&amp;nbsp;is monotonically decreasing.b) Show that the sequences&amp;nbsp;{an}&amp;nbsp;and&amp;nbsp;{bn}&amp;nbsp;are bounded.c) Deduce that the two sequences converge and prove that they converge to the same limit.

horieblersee275 · Accepted Answer

Given a1,b1 be two real numbers with 0&amp;lt;a1&amp;lt;b1 For n≥1, we definean+1=(anbn)12 and bn+1=an+bn21) Now,bn+1=an+bn2≥(anbn)12=an+1 or bn+1≥an+1 for all n.Thus, it implies thatan+1=(anbn)12≥(anan)12=an or, an+1≥anbn=bn+bn2≥an+bn2=bn+1 or, bn≥bn+1Hence, we proved that ∀n∈N, {an} is a monotonically increasing sequence and {bn} is a monotonically decreasing sequence. 2)Using 0&amp;lt;a1&amp;lt;b1, we have:b1≥bn≥bn+1≥an+1≥an≥a1&amp;gt;0 which implies that {an} is a bounded sequence bounded above by b1 and {bn} is a bounded sequence bounded below by b1.Monotonically increasing sequence is convergent and monotonically decreasing sequence is convergent which implies that both {an} and {bn} are convergent sequences.Let {an} converges to a and {bn} converges to b then,  b=a+b2 which gives us a=bThus, both sequences are convergent to the same limit.

Let \(\displaystyle{a}_{{1}},{b}_{{1}}\) be two real numbers such

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