91Ó°ÊÓ

consider the statement :"If $0≤f\left(x\right)≤g\left(x\right)$for every x$â‰\yen$0 and the improper integral ${∫}_0^{∞}g\left(x\right)dx$converges, then the improper integral ${∫}_0^{∞}f\left(x\right)dx$converges."

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

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

${∫}_0^{∞}g\left(x\right)dx$=A, Where A is finite.

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

Therefore,

$0≤{∫}_o^{∞}f\left(x\right)dx≤{∫}_0^{∞}g\left(x\right)dx$

${∫}_0^{∞}f\left(x\right)dx$=A

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

Therefore, the above statement is True.

consider the statement : "If $0≤f\left(x\right)≤g\left(x\right)$for every x>0; and $\underset{x→∞}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=3$, then the improper integrals ${∫}_0^{∞}g\left(x\right)dx$and${∫}_0^{∞}f\left(x\right)dx$ 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 $\underset{x→∞}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$:

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

The answer is 3.

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

Therefore, the above statement is False.

consider the statement: "If $0≤a_k\frac{1}{k}$for every positive integer k, then the series $\underset{k=1}{\overset{∞}{∑}}{a}_k$converges."

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

Use the comparison test.

The comparison test states that for $\underset{k=1}{\overset{∞}{∑}}{a}_k$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 $\underset{k=1}{\overset{∞}{∑}}{a}_k$converges.

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

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

Hence the above statement is False.

Consider the statement: If $\frac{1}{k^2}<b_k$for every integer k, then the series $\underset{k=1}{\overset{∞}{∑}}{b}_k$diverges.

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

Use the comparison test.

The comparison test states that for $\underset{k=1}{\overset{∞}{∑}}{a}_k$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 $\underset{k=1}{\overset{∞}{∑}}{a}_k$converges.

The comparison test fails to determine the divergence or convergence of the series $\underset{k=1}{\overset{∞}{∑}}{b}_k$.

Nothing can be said about the behavior of the series $\underset{k=1}{\overset{∞}{∑}}{b}_k$if $\frac{1}{k^2}=b_k$holds.

Hence the above statement is False.

Consider the statement: "If $a_k≤b_k$for every positive integer k and the series $\underset{k=1}{\overset{∞}{∑}}{b}_k$converges, then the series $\underset{k=1}{\overset{∞}{∑}}{a}_k$converges.

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

Consider the series $\underset{k=1}{\overset{∞}{∑}}{b}_k=\frac{1}{k^2}$and $\underset{k=1}{\overset{∞}{∑}}{a}_k=-\frac{1}{k}$.

Clearly, $a_k≤b_k$holds as:

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

The series $\underset{k=1}{\overset{∞}{∑}}{b}_k$=$\frac{1}{k^2}$is convergent by P-series testand the series $\underset{k=1}{\overset{∞}{∑}}{a}_k=-\frac{1}{k}$is divergent is P-series test.

Therefore , if $a_k≤b_k$for every positive integer k andthe series $\underset{k=1}{\overset{∞}{∑}}{b}_k$converges, then the series $\underset{k=1}{\overset{∞}{∑}}{a}_k$ converges is false.

Hence the above statement is False.

Consider the statement: " If the series $\underset{k=1}{\overset{∞}{∑}}{b}_k$and $\underset{k=1}{\overset{∞}{∑}}{a}_k$both diverge, then $\underset{k=1}{\overset{∞}{∑}}(a_k.b_k)$diverge.

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

Consider the series $\underset{k=1}{\overset{∞}{∑}}{b}_k=\frac{1}{k}$and $\underset{k=1}{\overset{∞}{∑}}{a}_k=\frac{1}{k}$.

The series $\underset{k=1}{\overset{∞}{∑}}{b}_k=\frac{1}{k}$the series $\underset{k=1}{\overset{∞}{∑}}{a}_k=\frac{1}{k}$are divergent by P-series test.

The series $\underset{k=1}{\overset{∞}{∑}}(a_k.b_k)=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k^2}$is convergent by p-series test.

Therefore, if the series $\underset{k=1}{\overset{∞}{∑}}{b}_k$and $\underset{k=1}{\overset{∞}{∑}}{a}_k$both diverge, then $\underset{k=1}{\overset{∞}{∑}}(a_k.b_k)$diverge is not true.

Hence the above statement is False.

Consider the statement: If $a_k$and $b_k$are both positive for every positive integer k and $\underset{k→∞}{\lim }\frac{a_k}{b_k}=\frac{1}{2}$, then $\underset{k=1}{\overset{∞}{∑}}{b}_k$and $\underset{k=1}{\overset{∞}{∑}}{a}_k$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 $\underset{k→∞}{\lim }\frac{a_k}{b_k}$is:

$\underset{k→∞}{\lim }\frac{a_k}{b_k}=\underset{k→∞}{\lim }\frac{1}{2}$

$=\frac{1}{2}$

But the series $\underset{k=1}{\overset{∞}{∑}}{a}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k}$and $\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k}$both diverge.

Hence, the given statement is not true.

Therefore, the above statement is False.

Consider the statement: "If $\underset{k=1}{\overset{∞}{∑}}{b}_k$and $\underset{k=1}{\overset{∞}{∑}}{a}_k$both converge, then $\underset{k→∞}{\lim }\frac{a_k}{b_k}$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 $\underset{k=1}{\overset{∞}{∑}}{a}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k^2}$and $\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k3}$both converge by P-series test.

The value of $\underset{k→∞}{\lim }\frac{a_k}{b_k}$is:

$\underset{k→∞}{\lim }\frac{a_k}{b_k}=\underset{k→∞}{\lim }\frac{k^3}{k^2}$

=$\underset{k→∞}{\lim }k$

=$∞$

Both the series$\underset{k=1}{\overset{∞}{∑}}{a}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k^2}$and $\underset{k=1}{\overset{∞}{∑}}{b}_k=\underset{k=1}{\overset{∞}{∑}}\frac{1}{k^3}$both converge but limit is not finite.

Hence, the given statement is not true.

Therefore, the above statement is False.