Given-u-0-a-u-n-1-a-1-2-1-1-b-n-u-n-a-gt-0-b-gt-1-a-Calculate-u-1-u-2-and-u-3-b-Show-that-the-sequence-u-n-n-N-is-increasing-c-Deduce-that-u-

Question Number 96489 by Ar Brandon last updated on 01/Jun/20

G iven {\begin{cases} u_{0} = a \\ u_{n + 1} = a + \frac{1}{2} (1 - \frac{1}{b^{n}}) u_{n} \end{cases} a > 0 b > 1

a ∖ C alculate u_{1}, u_{2}, and u_{3} .

b ∖ S how that the sequence {(u_{n})}_{n \in N} is increasing .

c ∖ D educe that {(u_{n})}_{n \in N} converges and determine its limit .

Answered by Rio Michael last updated on 01/Jun/20

(a) u_{0} = a \Rightarrow u_{1} = a + \frac{1}{2} (1 - \frac{1}{b^{0}}) u_{0}

\Rightarrow u_{1} = a

u_{2} = a + \frac{1}{2} (1 + \frac{1}{b}) a = a [1 + \frac{1}{2} (1 - \frac{1}{b})]

u_{3} = [a + \frac{1}{2} (1 - \frac{1}{b^{2}})] {a [1 + \frac{1}{2} (1 - \frac{1}{b})]}

(b) n \in N and a > 0, b > 1

a [1 + \frac{1}{2} (1 - \frac{1}{b})] > a for a > 0 and b > 1

[a + \frac{1}{2} (1 - \frac{1}{b^{2}})] {a [1 + \frac{1}{2} (1 - \frac{1}{b})]} > a [1 + \frac{1}{2} (1 - \frac{1}{b})]

\Rightarrow the sequence is increasing for n \in N .

(c) \frac{1}{2} (1 - \frac{1}{b^{n}}) u_{n} < 1 \forall n \in N \Rightarrow u_{n} is convergent

Commented by Ar Brandon last updated on 01/Jun/20

Wow, so fast ! Thank you ��