Q.1 - Calculus

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 ${\int }_0^{\infty }g\left(x\right)dx$ 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 $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=3$, 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.

Step-by-Step Solution

Answer

a) True

b) False

c) False

d) False

e) False

f) False 

g) False 

h) False 

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:

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

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.