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Determine whether the sequence is monotonic or eventually monotonic and whether the sequence is bounded above and/or below. If the sequence converges, give the limit.

135(2k1)147(3k2)

Short Answer

Expert verified

Ans: The sequence135(2k1)147(3k2)is convergent and its limit is0.

Step by step solution

01

Step 1. Given information.

given,

135(2k1)147(3k2)

02

Step 2. The objective is to determine whether the sequence is monotonic, bounded above, or bounded below, and to find the limit of the sequence if the sequence is convergent.  

The sequence {ak}=135(2k1)147(3k2)the general term is ak=135(2k1)147(3k2).

03

Step 3. The general term of the sequence is ak=1⋅3⋅5⋯(2k−1)1⋅4⋅7⋯(3k−2)

The ratio ak+1akgives

ak+1ak=135(2k1)(2k+1)147(3k2)(3k+1)135(2k1)147(3k2)(Substitution)=135(2k1)(2k+1)147(3k2)(3k+1)147(3k2)135(2k1)=2k+13k+1(Simplify)<1(Fork>0)

Thus, ak+1<akwhen the value of k>0.

The sequence {ak}=135(2k1)147(3k2)is eventually strictly increasing. The given sequence is monotonic.

04

Step 4. Now,

The sequence {ak}=135(2k1)147(3k2)is bounded below because

0<akfor k>0

As the index k increases, the term role="math" localid="1649275385618" {ak}=135(2k1)147(3k2)approaches 0.

Thus, the strictly decreasing sequence has an upper bound 1.

The given sequence has lower and upper bounds, therefore, the sequence is bounded.

05

Step 5. The monotonic increasing sequence is bounded above is convergent  

The strictly decreasing sequence {ak}=135(2k1)147(3k2)is bounded below and hence is convergent. Therefore, the sequence is convergent.

06

Step 6.  The limit of the sequence is {ak}=1⋅3⋅5⋯(2k−1)1⋅4⋅7⋯(3k−2)

limkak=limk135(2k1)147(3k2)=0(Simplify)

Thus, the limit of the sequence {ak}=135(2k1)147(3k2)is 0.

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