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Consider the sequence akdefined recursively bya1=1and fork>1,ak=2ak-1.Prove thatak2by first proving that the limit must exist. (Hint: Use induction to show that the terms of the sequence may be expressed with the closed formula forak=21-12k-1)

Short Answer

Expert verified

The value oflimkak=2

Step by step solution

01

Step 1. Given information

Consider the given sequence akdefined recursively bya1=1a nd fork>1,ak=2ak-1

02

Step 2. Finding the values of  terms a1,a2,a3

The terms of the sequence is defined as

a1=1.............(1)a2=2a1(putk=2inak=2ak-1)a2=2(simplify).................(2)

From equations,(1) and (2) ,it is observed that

0<a1<a2<2(because2<2)

Put k=3in ak=2ak-1to get

=22(substitutea2=2)

Thusa2<a3<2(because22<2)

03

Step 3. Finding the given sequence is bounded or not.

The general term of the sequence is defined as

ak=222.......(k-1)times

countinuing likewisw,the following inequality is obtained

0<a1<a2<a3<.......<ak<2

The sequenceak is an increasing sequence because

0<a1<a2<a3<.......<ak

the sequence akhas a lower bound and upper bound.Thus,the given sequence is bounded

04

Step 4. Find the value of limk→∞ak

The monotonic increasing sequence which is bounded above is convergent.The given sequence akdefined recursively by a1=1and for k>1,ak=2ak-1is increasing is bounded above by 2. thus,the sequence is convergent.

Assume the limit of the sequence akisl.

Therefore,

limkak=limk2ak-1l=2limkak-2(becauseak1)l=2ll2-2l=0(squaring)l(l-2)=0(factorize)l=0,2(solveforl)

The sequence cannot converge to 0 because a1=1 and the sequence is increasing.Therefore,the value of lis 2

Thus, the value of limkak=2

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