91Ó°ÊÓ

The term of series $\underset{k=1}{\overset{∞}{∑}}\frac{\sqrt{k+1}}{k^2+k+1}$is positive

The series of$\underset{k=1}{\overset{∞}{∑}}{b}_k$for the series $\underset{k=1}{\overset{∞}{∑}}\frac{\sqrt{k+1}}{k^2+k+1}$is given by

By dominating the series$\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{\sqrt{k}}{k^2}$

=$\underset{k=1}{\overset{∞}{∑}}\frac{1}{k^{3/2}}$.

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

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

The series $\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k^{3/2}}$is convergent by the p series.

Then the series $\underset{k=1}{\overset{∞}{∑}}{a}_k$is also convergent.

Therefore the series $\underset{k=1}{\overset{∞}{∑}}\frac{\sqrt{k+2}}{k^2+k+1}$is convergent and the

p-series is$\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k^{3/2}}$

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

It states that $\underset{k=1}{\overset{∞}{∑}}{a}_k$and $\underset{k=1}{\overset{∞}{∑}}{b}_k$be two series with positive terms such that $0≤a_k≤b_k$for every positive integer k.

If the series $\underset{k=1}{\overset{∞}{∑}}{b}_k$converges then the series $\underset{k=1}{\overset{∞}{∑}}{a}_k$also convergences

The term of series $\underset{k=1}{\overset{∞}{∑}}\sin \left(\frac{1}{k}\right)$are positive

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

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

The series $\underset{k=1}{\overset{∞}{∑}}{b}_k$for the series $\underset{k=1}{\overset{∞}{∑}}\sin \left(\frac{1}{k}\right)$is given by

$\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k}$

The series $\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k}$is divergent by the p- series

Therefore the $\underset{k=1}{\overset{∞}{∑}}{a}_k$is also divergent

Hence fore the $\underset{k=1}{\overset{∞}{∑}}\sin \left(\frac{1}{k}\right)$is divergent and p-series is$\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k}$

Consider the series $\underset{k=1}{\overset{∞}{∑}}\frac{k-1}{k^3+k+1}$

To determine p series that used to show that $\underset{k=1}{\overset{∞}{∑}}\frac{k-1}{k^3+k+1}$is convergent

The terms of series $\underset{k=1}{\overset{∞}{∑}}\frac{k-1}{k^3+k+1}$are positive.

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

$\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k^{3/2}}$

The ratio $\underset{k→∞}{\lim }\frac{a_k}{b_k}$is given

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

The value 0f $\underset{k→∞}{\lim }\frac{a_k}{b_k}=0$

The series $\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k^{3/2}}$is convergent by the p-series test

Then $\underset{k=1}{\overset{∞}{∑}}{a}_k$is also convergent

Then the series $\underset{k=1}{\overset{∞}{∑}}\frac{k-1}{k^3+k+1}$is convergent and the p series is $\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k^{3/2}}$