Chapter 8.4, Problem 37E

a.

To determine

To calculate: Whether the given series show the ratio test inconclusive.

Answer to Problem 37E

The given series show the ratio test inconclusive.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{n=\infty }{a}_n=}{\displaystyle \sum _{n=1}^{n=\infty }\frac{1}{n^3}}$

Concept Used:

Ratio Test:

(i) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L<1$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is absolutely convergent.

(and therefore convergent)

(ii) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L>1$ or $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\infty$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is divergent.

(iii) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=1$, the Ratio test is inconclusive.

Calculation:

The series is ${\displaystyle \sum _{n=1}^{n=\infty }{a}_n=}{\displaystyle \sum _{n=1}^{n=\infty }\frac{1}{n^3}}$

  $a_n=\frac{1}{n^3}$

Then,

  $a_{n+1}=\frac{1}{{\left(n+1\right)}^3}$

By use of ratio test

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{1}{{\left(n+1\right)}^3}\times \frac{n^3}{1}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{n^3}{{\left(n+1\right)}^3}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{1}{{\left(1+\frac{1}{n}\right)}^3}\right|\\ =\left|\frac{1}{{\left(1+\frac{1}{n}\right)}^3}\right|\to 1\text{as}\text{n}\to \infty \end{array}$

Here ,

  $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=1$ so, the Ratio test is inconclusive.

Conclusion:

The given series shows the ratio test inconclusive.

b.

To determine

To calculate:The given series show the ratio test inconclusive.

Answer to Problem 37E

The given series not shows the ratio test inconclusive.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }{a}_n}={\displaystyle \sum _{n=1}^{\infty }\frac{n}{2^n}}$

Calculation:

The series is ${\displaystyle \sum _{n=1}^{\infty }{a}_n}={\displaystyle \sum _{n=1}^{\infty }\frac{n}{2^n}}$

  $a_n=\frac{n}{2^n}$

Then,

  $a_{n+1}=\frac{n+1}{2^{n+1}}$

By use of ratio test

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{n+1}{2^{n+1}}\times \frac{2^n}{n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{n+1}{2n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{1+\frac{1}{n}}{2}\right|\\ =\left|\frac{1+\frac{1}{n}}{2}\right|\to \frac{1}{2}\text{as }\text{n}\to \infty \end{array}$

Here,

  $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{2}<1$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is absolutely convergent.

Conclusion:

The given series not shows the ratio test inconclusive.

c.

To determine

To calculate:The given series show the ratio test inconclusive.

Answer to Problem 37E

The given series not shows the ratio test inconclusive.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{n=\infty }\frac{{\left(-3\right)}^{n-1}}{\sqrt{n}}}$

Concept Used:

Ratio Test:

(iv) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L<1$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is absolutely convergent.

(and therefore convergent)

(v) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L>1$ or $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\infty$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is divergent.

(vi) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=1$, the Ratio test is inconclusive.

Calculation:

The series is ${\displaystyle \sum _{n=1}^{\infty }{a}_n}={\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-3\right)}^{n-1}}{\sqrt{n}}}$

  $a_n=\frac{{\left(-3\right)}^{n-1}}{\sqrt{n}}$

Then,

  $a_{n+1}=\frac{{\left(-3\right)}^n}{\sqrt{\left(n+1\right)}}$

By use of ratio test

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{{\left(-3\right)}^n}{\sqrt{\left(n+1\right)}}\times \frac{\sqrt{n}}{{\left(-3\right)}^{n-1}}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{\left(-3\right)\sqrt{n}}{\sqrt{\left(n+1\right)}}\right|\\ =\underset{n\to \infty }{\lim }\left|-3\right|\frac{\sqrt{n}}{\sqrt{\left(n+1\right)}}\\ =\underset{n\to \infty }{\lim }3\cdot \frac{1}{\sqrt{1+\frac{1}{n}}}\\ =3\cdot \frac{1}{\sqrt{1+\frac{1}{n}}}\to 3\text{as}\text{n}\to \infty \\ \end{array}$

Here ,

  $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=3>1$ then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is divergent.

Conclusion:

The given series not shows the ratio test inconclusive.

d.

To determine

To calculate:The given series show the ratio test inconclusive.

Answer to Problem 37E

The given series shows the ratio test inconclusive.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{n=\infty }\frac{\sqrt{n}}{1+n^2}}$

Concept Used:

Ratio Test:

(vii) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L<1$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is absolutely convergent.

(and therefore convergent)

(viii) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L>1$ or $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\infty$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is divergent.

(ix) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=1$, the Ratio test is inconclusive.

Calculation:

The series is ${\displaystyle \sum _{n=1}^{n=\infty }{a}_n=}{\displaystyle \sum _{n=1}^{n=\infty }\frac{\sqrt{n}}{1+n^2}}$

  $a_n=\frac{\sqrt{n}}{1+n^2}$

Then,

  $a_{n+1}=\frac{\sqrt{\left(n+1\right)}}{1+{\left(1+n\right)}^2}$

By use of ratio test

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{\sqrt{n+1}}{1+{\left(1+n\right)}^2}\times \frac{1+n^2}{\sqrt{n}}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{\sqrt{n+1}}{\sqrt{n}}\times \frac{1+n^2}{2+n^2+2n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{\sqrt{1+\frac{1}{n}}}{1}\times \frac{\frac{1}{n^2}+1}{\frac{2}{n^2}+1+\frac{2}{n}}\right|\to 1\\ \left|\frac{a_{n+1}}{a_n}\right|\to 1\text{as}\text{n}\to \infty \end{array}$

Here ,

  $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=1$ then the Ratio test is inconclusive.

Conclusion:

The given series shows the ratio test inconclusive.