Chapter 9.5, Problem 58E

a.

To determine

To find: To find the alternating series test.

Answer to Problem 58E

The alternating series test value is $a_k={(-1)}^{k+1}\left(\frac{2}{k}\right)$ .

Explanation of Solution

Given:

The given series $a_k={(-1)}^{k+1}{\displaystyle {\int }_0^{\frac{1}{k}}6{(kx)}^{2}dx}$

Calculation:

Given series is $a_k={(-1)}^{k+1}{\displaystyle {\int }_0^{\frac{1}{k}}6{(kx)}^{2}dx}$ .

  $\begin{array}{}$ a_k={(-1)}^{k+1}{\displaystyle {\int }_0^{\frac{1}{k}}6{(kx)}^{2}dx}$$ a_k={(-1)}^{k+1}{[2k^2{x}^3]}_0^{\frac{1}{k}}$$ a_k={(-1)}^{k+1}\left(\frac{2}{k}\right)$\end{array}$

b.

To determine

To find: To find whether the given series converges or diverges.

Answer to Problem 58E

The series converges.

Explanation of Solution

Given:

The given series $a_k={(-1)}^{k+1}{\displaystyle {\int }_0^{\frac{1}{k}}6{(kx)}^{2}dx}$ .

Formula used:

  $a=f(n)$where $f$ is a continuous, positive, decreasing function of x for all $x\ge N$ (N is a positive integer).

Then the series ${\displaystyle \sum _{n=N}^{\infty }{a}_n}$ and the integral ${\displaystyle {\int }_N^{\infty }f(x)dx}$ either both converges or both diverges.

Calculation:

Given series $a_k={(-1)}^{k+1}{\displaystyle {\int }_0^{\frac{1}{k}}6{(kx)}^{2}dx}$ .

The series converges by the alternating series test.

c.

To determine

To find interval of sum.

Answer to Problem 58E

The interval of the sum is $1\le {\displaystyle \sum _{k=1}^{\infty }{a}_k}\le \frac{3}{2}$ .

Explanation of Solution

Given:

The given series $a_k={(-1)}^{k+1}{\displaystyle {\int }_0^{\frac{1}{k}}6{(kx)}^{2}dx}$

Formula used:

Sum is given for an infinite geometric progression is given by $S=\frac{a}{1-r}$ where a is first and r is common ratio, also $-1<r<1$ .

Calculation:

Given series is $a_k={(-1)}^{k+1}{\displaystyle {\int }_0^{\frac{1}{k}}6{(kx)}^{2}dx}$ .

  $\begin{array}{l}{S}_1=2\\ S_2=1\\ S_3=\frac{5}{3}\\ S_4=\frac{7}{6}\\ S_5=\frac{47}{30}\\ S_6=\frac{37}{30}\\ S_7=\frac{319}{210}\\ S_8=\frac{533}{420}\\ S_9=\frac{1879}{1260}\end{array}$

For alternating series, the sum is between any two adjacent partial sums, so,

  $\begin{array}{l}1<S_8\le sum\le S_9<\frac{3}{2}\\ 1\le {\displaystyle \sum _{k=1}^{\infty }{a}_k}\le \frac{3}{2}\end{array}$