Chapter 8, Problem 1RE

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

1. a. The terms of the sequence {a_n} increase in magnitude, so the limit of the sequence does not exist.
2. b. The terms of the series ${\displaystyle \sum 1/\sqrt{k}}$ approach zero, so the series converges.
3. c. The terms of the sequence of partial sums of the series approach so the infinite series ∑a_k approach $\frac{5}{2}$, so the infinite series converges to $\frac{5}{2}$.
4. d. An alternating series that converges absolutely must converge conditionally.
5. e. The sequence $a_n=\frac{n^2}{n^2+1}$ converges.
6. f. The sequence $a_n=\frac{{(-1)}^n{n}^2}{n^2+1}$ converges.
7. g. The series ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{k^2}{k^2+1}}$ converges.
8. h. The sequence of partial sums associated with the series ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{k^2}{k^2+1}}$ converges.

To determine

Whether the statement “The terms of the sequence $\left\{a_n\right\}$ increase in magnitude, so the limit of the sequence does not exist” is true or not.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The n the term of the sequence is given as $a_n=2-\frac{1}{n^2}$.

Obtain the limit of $a_n=2-\frac{1}{n^2}$.

$\begin{array}{c}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left(2-\frac{1}{n^2}\right)\\ =\underset{n\to \infty }{\lim }\left(2\right)-\underset{n\to \infty }{\lim }\left(\frac{1}{n^2}\right)\\ =2-\frac{1}{\infty }\\ =2\end{array}$

That is, the limit exists.

Hence, the given statement is false.

To determine

Whether the statement “The terms of the series ${\displaystyle \sum \frac{1}{\sqrt{k}}}$ approach zero, so the series converges” is true or not.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Result used:

If the series converges, then $\underset{k\to \infty }{\lim }{a}_k=0$. Converse need not be true.

Calculation:

Consider k th term of the series as $a_k=\frac{1}{k}$.

Obtain the limit of $a_k$.

$\begin{array}{c}\underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\frac{1}{k}\\ =\frac{1}{\infty }\\ =0\end{array}$

Thus, by the result stated above note that the terms of the sequence tend to zero is not sufficient for convergence of the series.

Hence, the given statement is false.

To determine

Whether the statement “The terms of the sequence of partial sums of the series ${\displaystyle \sum a_k}$ approach $\frac{5}{2}$ , so the infinite series converges to $\frac{5}{2}$” is true or not.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Definition used:

If the sequence of partial sums $\left\{S_n\right\}$ has a limit $L$, the infinite series converges to that limit.

Calculation:

From the given statement, the terms of the sequence of partial sums of the series ${\displaystyle \sum a_k}$ approach $\frac{5}{2}$.

By the definition stated above, the infinite series converges to $\frac{5}{2}$.

Hence, the given statement is true.

To determine

Whether the statement “An alternating series that converges absolutely must converge conditionally” is true or not.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Definition used:

(1) The series ${\displaystyle \sum a_k}$ converges absolutely if ${\displaystyle \sum a_k}\text{ and }{\displaystyle \sum \left|a_k\right|}$ converges.

(2) The series ${\displaystyle \sum a_k}$ converges conditionally if ${\displaystyle \sum a_k}$ converges but ${\displaystyle \sum \left|a_k\right|}$ diverges.

Calculation:

From the given statement, the series converges absolutely.

By the definition (1) stated above, the series ${\displaystyle \sum a_k}\text{ and }{\displaystyle \sum \left|a_k\right|}$ both are converges.

Since the absolute value of the ${\displaystyle \sum \left|a_k\right|}$ converges and by the definition (2) stated above, it can be concluded that the series does not conditionally converge.

Hence, the given statement is false.

To determine

Whether the statement “The sequence $a_n=\frac{n^2}{n^2+1}$ converges” is true or not.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Result used:

If the sequence has the limit, then the sequence converges.

Calculation:

Obtain the limit of $a_n$.

$\begin{array}{c}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{n^2}{n^2+1}\\ =\underset{n\to \infty }{\lim }\frac{n^2}{n^2\left(1+\frac{1}{n^2}\right)}\\ =\frac{1}{\left(1+\frac{1}{\infty }\right)}\\ =1\end{array}$

Since the limit of the sequence exist and result stated above, it can be concluded that the sequence converges.

Hence, the given statement is true.

To determine

Whether the statement “The sequence $a_n=\frac{{\left(-1\right)}^n{n}^2}{n^2+1}$ converges” is true or not.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Result used:

If the sequence has the limit, then the sequence converges otherwise it is diverges.

Calculation:

Obtain the limit of $a_n$.

$\begin{array}{c}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{{\left(-1\right)}^n{n}^2}{n^2+1}\\ =\underset{n\to \infty }{\lim }\frac{{\left(-1\right)}^n{n}^2}{n^2\left(1+\frac{1}{n^2}\right)}\\ =\frac{{\left(-1\right)}^n}{\left(1+\frac{1}{\infty }\right)}\\ ={\left(-1\right)}^n\end{array}$

If $n$ is odd, the sequence has limit $-1$ and if n is even the sequence has limit 1.

That is, the sequence does not have the same limit.

Thus, by result stated above, it can be concluded that the sequence does not converge.

Hence, the given statement is false.

To determine

Whether the statement “The series ${\displaystyle \sum _{k=1}^{\infty }\frac{k^2}{k^2+1}}$ converges” is true or not.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Result used:

If $\underset{k\to \infty }{\lim }{a}_k\ne 0$, then the series diverges.

Calculation:

Obtain the limit of $a_k=\frac{k^2}{k^2+1}$.

$\begin{array}{c}\underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\frac{k^2}{k^2+1}\\ =\underset{k\to \infty }{\lim }\frac{k^2}{k^2\left(1+\frac{1}{k^2}\right)}\\ =\frac{1}{\left(1+\frac{1}{\infty }\right)}\\ =1\end{array}$

Since $\underset{k\to \infty }{\lim }{a}_k\ne 0$ and by result stated above, it can be concluded that the series diverges.

Hence, the given statement is false.

To determine

Whether the statement “The sequence of partial sums association with the series ${\displaystyle \sum _{k=1}^{\infty }\frac{1}{k^2+1}}$ converges” is true or not.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Result used:

(1) “Let ${\displaystyle \sum a_k}$ and ${\displaystyle \sum b_k}$ be series with positive terms.

1. (a) If $0<a_k\le b_k$ and ${\displaystyle \sum b_k}$ converges, then ${\displaystyle \sum a_k}$ converges.
2. (b) If $0<b_k\le a_k$ and ${\displaystyle \sum b_k}$ diverges, then ${\displaystyle \sum a_k}$ diverges.

(2) The p-series ${\displaystyle \sum _{k=1}^{\infty }\frac{1}{k^p}}$ is convergent if $p>1$ and diverges if $p\le 1$.

Calculation:

Let $k^2<k^2+1$

$\begin{array}{l}\frac{1}{k^2+1}<\frac{1}{k^2}\\ {\displaystyle \sum _{k=1}^{\infty }\frac{1}{k^2+1}<}{\displaystyle \sum _{k=1}^{\infty }\frac{1}{k^2}}\end{array}$

In the series ${\displaystyle \sum _{k=1}^{\infty }\frac{1}{k^2}}$, $p=2$. That is, the value of p is greater than 1.

Thus, by the Result (2), the series ${\displaystyle \sum _{k=1}^{\infty }\frac{1}{k^2}}$ is convergent.

Therefore, by the Result (1), it can be concluded that the given series convergent.

Hence, the series ${\displaystyle \sum _{k=1}^{\infty }\frac{1}{k^2+1}}$ converges.

Hence, the given statement is true.