Q.2 - Calculus

Question

a) As p- series you could use as a comparison to show that the series $\sum _{k=1}^{\infty }\frac{\sqrt{k+2}}{k^2+k+1}$converges.

b) As a p- series you could use a comparison to show that the series  $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k}\right)$diverges.

c) A  p series other than $\sum _{k=1}^{\infty }\frac{1}{k^2}$you could use with comparison test to show that the series $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$ converges.

Step-by-Step Solution

Answer

a) It is convergent

b) It is divergent

c) it is convergent

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

The term of series $\sum _{k=1}^{\infty }\frac{\sqrt{k+1 }}{k^2+k+1}$is positive

The series of$\sum _{k=1}^{\infty }{b}_k$ for the series $\sum _{k=1}^{\infty }\frac{\sqrt{k+1}}{k^2+k+1}$is given by

 By dominating the series$\sum _{k=1 }^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{\sqrt{k}}{k^2}$

=$\sum _{k=1}^{\infty }\frac{1}{k^{3/2}}$.

a) step2 :The ratio of limit is given by lim k → ∞ a k b k :

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{{\displaystyle \frac{\sqrt{k+2}}{k^2+k+1}}}{{\displaystyle \frac{1}{k^{3/2}}}} \\ = \underset{k\to \infty }{\lim }\frac{k^{3/2}\sqrt{k+2}}{k^2+k+1} \\ = \underset{k\to \infty }{\lim }\frac{k^2\sqrt{1+{\displaystyle \frac{2}{k}}}}{k^2(1+{\displaystyle \frac{1}{k}}+{\displaystyle \frac{1}{k^2}})} \\ =\underset{k\to \infty }{\lim }\frac{\sqrt{1+{\displaystyle \frac{2}{k}}}}{1+{\displaystyle \frac{1}{k}}+{\displaystyle \frac{1}{K^2}}} \\ =1 \end{gathered}$$

a) step 3 Therefore by the solution

The value of$\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=1$ which is non- zero finite number.

The series $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{3/2}}$ is convergent by the p series.

Then the series $\sum _{k=1}^{\infty }{a}_k$ is also convergent.

Therefore the series $\sum _{k=1}^{\infty }\frac{\sqrt{k+2}}{k^2+k+1}$ is convergent and the

 p-series is$\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{3/2}}$

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

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

It states that $\sum _{k=1}^{\infty }{a}_k$and  $\sum _{k=1}^{\infty }{b}_k$be two series with positive terms such that $0\le a_k\le b_k$ for every positive integer k.

If the series $\sum _{k=1}^{\infty }{b}_k$converges then the series $\sum _{k=1}^{\infty }{a}_k$also convergences

b) step 2 According to the given data

The term of series $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k}\right)$ are positive

The expression of $\sin \left(\frac{1}{k}\right)$ follows inequality

$\sin \left(\frac{1}{k}\right)\le \frac{1}{k}$

The series $\sum _{k=1}^{\infty }{b}_k$for the series  $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k}\right)$is given by

$\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k}$

b) step 3 By concluding

The series $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k}$ is divergent by the p- series

Therefore the $\sum _{k=1}^{\infty }{a}_k$is  also divergent

Hence fore the $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k}\right)$ is divergent and p-series is$\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k}$

c)

Consider the series $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$

To determine p series that used to show that $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$ is convergent

The terms of series  $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$are positive.

The series $\underset{k=1}{\overset{\infty }{\sum b_k}}$ for the series $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$is

$\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{3/2}}$

c) step 2

The ratio $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}$is given

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{{\displaystyle \frac{k-1}{k^3+k+1}}}{{\displaystyle \frac{1}{k^{3/2}}}} \\ =\underset{k\to \infty }{\lim }\frac{k^{3/2}(k-1)}{k^3+k+1} \\ =\underset{k\to \infty }{\lim }\frac{k^{5/2}(1+{\displaystyle \frac{1}{k}})}{k^3(1+{\displaystyle \frac{1}{k^2}}+{\displaystyle \frac{1}{k^3}})} \\ =\underset{k\to \infty }{\lim }\frac{(1+{\displaystyle \frac{1}{k}})}{k^{1/2}(1+{\displaystyle \frac{1}{k^2}}+{\displaystyle \frac{1}{k^3}})} \\ =0 \end{gathered}$$

c) step 3

The value 0f $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=0$

The series $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{3/2}}$ is convergent by the p-series test

Then $\sum _{k=1}^{\infty }{a}_k$ is also convergent

Then the series $\sum _{k=1}^{\infty }\frac{k-1}{k^3+k+1}$is convergent and the p series is $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^{3/2}}$