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a) As p- series you could use as a comparison to show that the series k=1k+2k2+k+1converges.

b) As a p- series you could use a comparison to show that the series k=1sin1kdiverges.

c) A p series other than k=11k2you could use with comparison test to show that the series k=1k-1k3+k+1converges.

Short Answer

Expert verified

a) It is convergent

b) It is divergent

c) it is convergent

Step by step solution

01

a)The objective is to find the p- series that is used to show that the series ∑k=1∞k+2k2+k+1is convergent

The term of series k=1k+1k2+k+1is positive

The series ofk=1bkfor the series k=1k+1k2+k+1is given by

By dominating the seriesk=1bk=k=1kk2

=k=11k3/2.

02

a) step2 :The ratio of limit is given by limk→∞akbk:

limkakbk=limkk+2k2+k+11k3/2=limkk3/2k+2k2+k+1=limkk21+2kk2(1+1k+1k2)=limk1+2k1+1k+1K2=1

03

a) step 3 Therefore by the solution

The value oflimkakbk=1which is non- zero finite number.

The series k=1bk=k=11k3/2is convergent by the p series.

Then the series k=1akis also convergent.

Therefore the series k=1k+2k2+k+1is convergent and the

p-series isk=1bk=k=11k3/2

04

b)The objective is to find the p- series that is used to show that the series ∑k=1∞sin1k is convergent

The comparison test is used to determine the convergence or divergence of the series

It states that k=1akand k=1bkbe two series with positive terms such that 0akbkfor every positive integer k.

If the series k=1bkconverges then the series k=1akalso convergences

05

b) step 2 According to the given data

The term of series k=1sin1kare positive

The expression of sin1kfollows inequality

sin1k1k

The series k=1bkfor the series k=1sin1kis given by

k=1bk=k=11k

06

b) step 3 By concluding

The series k=1bk=k=11kis divergent by the p- series

Therefore the k=1akis also divergent

Hence fore the k=1sin1kis divergent and p-series isk=1bk=k=11k

07

c) 

Consider the series k=1k-1k3+k+1

To determine p series that used to show that k=1k-1k3+k+1is convergent

The terms of series k=1k-1k3+k+1are positive.

The series bkk=1for the series k=1k-1k3+k+1is

k=1bk=k=11k3/2

08

c) step 2

The ratio limkakbkis given

limkakbk=limkk-1k3+k+11k3/2=limkk3/2(k-1)k3+k+1=limkk5/2(1+1k)k3(1+1k2+1k3)=limk(1+1k)k1/2(1+1k2+1k3)=0

09

c) step 3

The value 0f limkakbk=0

The series k=1bk=k=11k3/2is convergent by the p-series test

Then k=1akis also convergent

Then the series k=1k-1k3+k+1is convergent and the p series is k=1bk=k=11k3/2

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