Question

True /False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(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="1651646409965" ${\int }_0^{\infty }g\left(x\right)dx$converges, then the improper integral ${\int }_0^{\infty }f\left(x\right)dx$.

(b) True or False: If $0\le f\left(x\right)\le g\left(x\right)$for every x>0 and role="math" localid="1651646831120" $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=3$, then the improper integrals ${\int }_0^{\infty }f\left(x\right)dx$both converge.

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

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

(e) True or False: If $a_k\le b_k$for every positive integer k and the series $\sum _{k=1}^{\infty }{b}_k$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$\sum _{k=1}^{\infty }{b}_k$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) step1

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 determine to 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 finite.

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

Therefore,

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

${\int }_0^{\infty }f\left(x\right)dx$=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 determine to 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 $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$:

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

The answer is 3.

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

Therefore, the above statement is False.

c) step1

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 determine to 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.

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) step1

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

The objective is to determine 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.

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 is False.

e) step1

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 determine to 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:

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

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

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

Hence the above statement is False.

f) step1

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 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.

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) step1

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 determine whether the statement is true or false.

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

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

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

$=\frac{1}{2}$

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 determine whether the statement is true or false.

Consider the functions $a_k=\frac{1}{k^2}$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 $\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k3}$both converge by P-series test.

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

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

=$\underset{k\to \infty }{\lim }k$

=$\infty$

Both 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 but limit is not finite.

Hence, the given statement is not true.

Therefore, the above statement is False.