Answered: Let x1 = 1 and define xk+1 = sqrt(2xk)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Let x₁= 1 and define x_k+1 = sqrt(2x_k) where k is a natural number. Prove that the sequence {x_k} for k = 1 to infinity converges and find its limit.

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Step 1

Given: $x_{1} = 1, x_{k + 1} = \sqrt{2 x_{k}}$

So, $x_{2} = \sqrt{2 x_{1}} = \sqrt{2} > x_{1}$

$x_{3} = \sqrt{2 x_{2}} = \sqrt{2 \sqrt{2}} > x_{2}$

On generalizing, we get

$x_{1} < x_{2} < x_{3} . . . . . < x_{k}$

So, sequence $\{x_{k}\}$ is an increasing sequence and hence monotonic

Also, $1 \leq x_{k} < 2 \forall k$

So, $\{x_{k}\}$ is bounded

As the sequence $\{x_{k}\}$ is monotonic as well as bounded . So, by Monotone Convergence Theorem it must be convergent.

That is , $\{x_{k}\}$ is convergent.

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