Question

(a) True or False. If $0\le f\left(x\right)\le g\left(x\right)$for every $x\ge 0$and the improper integral role="math" localid="1651759436770" ${\int }_0^{\infty }g\left(x\right)dx$converges, then the improper integral role="math" localid="1651759467269" ${\int }_0^{\infty }f\left(x\right)dx$converges.

(b) True or False. If $0\le f\left(x\right)\le g\left(x\right)$for every $x>0$and $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=3,$then the improper integrals${\int }_0^{\infty }g\left(x\right)dx$and ${\int }_0^{\infty }f\left(x\right)dx$both converge.

(c) True or False. If $0\le a_k<\frac{1}{k}$for every positive integer $k$, then the series $\sum _{k=1}^{\infty }{a}_k$converges.

(d) True or False. If $\frac{1}{k^2}<b_k$for every positive integer $k$, then the series role="math" localid="1651759979364" $\sum _{k=1}^{\infty }{b}_k$diverges.

(e) True or False. If $a_k\le b_k$for every positive integer $k$and the series role="math" localid="1651760977273" $\sum _{k=1}^{\infty }{b}_k$ converges, then the series $\sum _{k=1}^{\infty }{a}_k$converges.

(f) True or False. If $\sum _{k=1}^{\infty }{a}_k$and $\sum _{k=1}^{\infty }{b}_k$both diverge, then $\sum _{k=1}^{\infty }(a_k.b_k)$diverges.

(g) True or False. If $a_k$and $b_k$are both positive for every positive integer $k$and $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\frac{1}{2}$, then $\sum _{k=1}^{\infty }{a}_k$and $\sum _{k=1}^{\infty }{b}_k$both converge.

(h) True or False. If $\sum _{k=1}^{\infty }{a}_k$and $\underset{k=1}{\overset{\infty }{\sum b_k}}$both converge, then $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}$is finite.

Short answer

(a) True

(b) False

(c) False

(d) False

(e) False

(f) False

(g) False

(h) False

Step-by-step solution

(a) Step 1:

Consider the statement: "If $0\le f\left(x\right)\le g\left(x\right)$for every $x\ge 0$and the improper integral ${\int }_0^{\infty }g\left(x\right)dx$converges, then the improper integral ${\int }_0^{\infty }f\left(x\right)dx$converges."

The objective is to determine whether the statement is true or false.

The improper integral${\int }_0^{\infty }g\left(x\right)dx$is convergent. Therefore,

${\int }_0^{\infty }g\left(x\right)dx=A$,where A is a finite.

(a) Step 2:

It is given that$0\le f\left(x\right)\le g\left(x\right)$.

Therefore,

$0\le {\int }_0^{\infty }f\left(x\right)dx\le {\int }_0^{\infty }g\left(x\right)dx$

${\int }_0^{\infty }f\left(x\right)dx\le A$

Therefore, the improper integral ${\int }_0^{\infty }f\left(x\right)dx$converges.

Therefore, the above statement is TRUE.

 (b) Step 1:

Consider the statement: "If $0\le f\left(x\right)\le g\left(x\right)$for every $x>0$; and $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=3$, then the improper integrals ${\int }_0^{\infty }g\left(x\right)dx$and ${\int }_0^{\infty }f\left(x\right)dx$both converge."

The objective is to determine whether the statement is true or false

Consider the functions $g\left(x\right)=\frac{1}{x^2}$and$f\left(x\right)=\frac{3}{x^2}$.

The value of $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$is:

$\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim 3}$

$=3$

(b) Step 2:

But the integrals ${\int }_0^{\infty }g\left(x\right)dx={\int }_0^{\infty }\frac{1}{x^2}dx$and ${\int }_0^{\infty }f\left(x\right)dx={\int }_0^{\infty }\frac{3}{x^2}dx$diverges.

Therefore, the above statement is False.

(c) Step 1:

Consider the statement: "If$0\le a_k<\frac{1}{k}$for every positive integer$k$, then the series$\sum _{k=1}^{\infty }{a}_k$converges."

The objective is to determine whether the statement is true or false.

To determine whether the statement is true or false, use the comparison test.

The comparison test states that for $\sum _{k=1}^{\infty }{a}_k$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 $\sum _{k=1}^{\infty }{a}_k$converges.

  (c) Step 2:

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

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

Hence, the above statement is False.

 (d) Step 1:

Consider the statement: "If $\frac{1}{k^2}<b_k$for every positive integer $k$, then the series $\sum _{k=1}^{\infty }{b}_k$diverges."

The objective is to determine whether the statement is true or false.

To determine whether the statement is true or false, use the comparison test.

The comparison test states that for $\sum _{k=1}^{\infty }{a}_k$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 . If the series $\sum _{k=1}^{\infty }{b}_k$converges, then the series $\sum _{k=1}^{\infty }{a}_k$converges.

 (d) Step 2:

The comparison test fails to determine the divergence or convergence of the series $\sum _{k=1}^{\infty }{b}_k$.

Nothing can be said about the behavior of the series $\sum _{k=1}^{\infty }{b}_k$if $\frac{1}{k^2}<b_k$holds.

Hence, the above statement isFalse.

(e) Step 1:

Consider the statement: "If$a_k\le b_k$for every positive integer$k$and the series$\sum _{k=1}^{\infty }{b}_k$converges, then the series$\sum _{k=1}^{\infty }{a}_k$converges."

The objective is to determine whether the statement is true or false.

Consider the series$\sum _{k=1}^{\infty }{b}_k=\frac{1}{k^2}$ and$\sum _{k=1}^{\infty }{a}_k=-\frac{1}{k}$.

Clearly,$a_k\le b_k$holds as:

$-\frac{1}{k}<\frac{1}{k^2}$for$k>0$

 (e) Step 2:

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

Therefore, if $a_k\le b_k$for every positive integer $k$and the series $\sum _{k=1}^{\infty }{b}_k$converges, then the series $\sum _{k=1}^{\infty }{a}_k$converges is false.

Hence, the above statement isFalse.

(f) Step 1:

Consider the statement: "If the series$\sum _{k=1}^{\infty }{b}_k$and$\sum _{k=1}^{\infty }{a}_k$both diverge, then$\sum _{k=1}^{\infty }(a_k.b_k)$diverge."

The objective is to determine whether the statement is true or false.

Consider the series$\sum _{k=1}^{\infty }{b}_k=\frac{1}{k}$and$\sum _{k=1}^{\infty }{a}_k=\frac{1}{k}$.

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

(f) Step 2:

The series $\sum _{k=1}^{\infty }(a_k.b_k)=\sum _{k=1}^{\infty }\frac{1}{k^2}$is convergent by p-series test.

Therefore, if the series $\sum _{k=1}^{\infty }{b}_k$and $\sum _{k=1}^{\infty }{a}_k$both diverge, then $\sum _{k=1}^{\infty }(a_k.b_k)$diverge is not true.

Hence, the above statement is False.

(g) Step 1:

Consider the statement: "If$a_k$and$b_k$are both positive for every positive integer$k$and$\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\frac{1}{2}$, then$\sum _{k=1}^{\infty }{b}_k$and$\sum _{k=1}^{\infty }{a}_k$both converge."

The objective is to determine whether the statement is true or false.

Consider the functions$a_k=\frac{1}{k}$and$b_k=\frac{2}{k}$.

The value of $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}$is:

$\underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{1}{2}$

$=\frac{1}{2}$

(g) Step 2:

But the series$\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{1}{k}$and$\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k}$both diverge.

Hence, the given statement is not true.

Therefore, the above statement is False.

 (h) Step 1:

Consider the statement: "If $\sum _{k=1}^{\infty }{b}_k$and $\sum _{k=1}^{\infty }{a}_k$both converge, then $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}$is finite."

The objective is to determine whether the statement is true or false.

Consider the function $a_k=\frac{1}{k^2}$and $b_k=\frac{1}{k^3}$.

But the series $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{1}{k^2}$and $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^3}$both converge by p-series test.

 (h) Step 2:

The value of $\underset{k\to \infty }{\lim }\frac{a_k}{b_k}$is:

$$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_k}{b_k}=\underset{k\to \infty }{\lim }\frac{k^3}{k^2} \\ =\underset{k\to \infty }{\lim k} \\ =\infty \end{gathered}$$