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

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.1: Infinite Sequences And Summation Notation

Problem 33E

See similar textbooks

Related questions

Concept explainers

Topic Video

Question

100%

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.

Expert Solution

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.

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Sequence$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions