Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.5


Exercise 2.5.1

(a) Impossible by Theorem 2.5.5.

(b) Consider the sequence given by\[a_n=\frac{1+(-1)^n+\frac{1}{n}}{2}.\]Then we have\[a_{2n}=1+\frac{1}{4n},~a_{2n-1}=\frac{1}{2(2n-1)}.\]It is clear that $(a_{2n})$ converges to $1$ and $(a_{2n-1})$ converges to $0$. Moreover, it is easy to check that $a_n\ne 0$ and $a_n \ne 1$.

(c) Consider the sequence \begin{gather*}1,\\ 1,~\frac{1}{2},\\ 1,~\frac{1}{2},~\frac{1}{3},\\ \cdots \\ 1,~\frac{1}{2},~\frac{1}{3},\cdots,~ \frac{1}{n},\\ \cdots.\end{gather*}

(d) Impossible. Because there must exist a subsequence who converges to zero while zero is not in this set.


Exercise 2.5.2

(a) True. Consider the proper subsequence $(x_2,x_3,\dots,)$. If it is convergent, then the sequence $(x_n)$ is also convergent.

(b) True. Suppose $(x_n)$ converges, then by Theorem 2.5.2, all subsequences of $(x_n)$ are convergent. This contradicts the assumption. Hence $(x_n)$ diverges.

(c) True. Since $(x_n)$ is bounded, there exists a subsequence $(x_{n_k})$ who converges. Let $a$ be the limit of this subsequence. Since $(x_n)$ diverges, there exists $\delta>0$ such that for all $N\in\mathbf N$ there exists $n> N$ such that $|x_n-a|\geqslant \delta$. In particular, there exists a subsequence $(x_{m_k})$ such that $|x_{m_k}-a|\geqslant \delta$ for all $k\in\mathbf N$. As a subsequence of a bounded sequence $(x_n)$, $(x_{m_k})$ is also bounded and hence by Bolzano-Weierstrass Theorem it has a convergent subsequence $(x_{r_k})$. But because of $|x_{m_k}-a|\geqslant \delta$ for all $k\in\mathbf N$, by Order Limit Theorem, we also have $\left|\lim x_{r_k}- a\right|\geqslant \delta$. Therefore, the two subsequences $(x_{n_k})$ and $(x_{r_k})$ converge to different limits.

(d) True. For convenience, we assume that $(x_n)$ is increasing and the subsequence $(x_{n_k})$ converges. Let $a$ be the limit of he subsequence $(x_{n_k})$. Since it is increasing, we have $x_{n_k}\leqslant a$. Moreover for any $n\in\mathbf N$, we can find $k$ such that $n_k> n$, then\[x_n\leqslant x_{n_k} \leqslant a.\]Now we show that $\lim x_n=a$.

Let $\varepsilon>0$ be arbitrary. Since $\lim x_{n_k}=a$, there exists $m$ such that for all $k>m$, we have \[|x_{n_k}-a|<\varepsilon.\]As $x_{n_k}\leqslant a$, we have\[-\varepsilon < x_{n_k}-a\leqslant 0.\]Then for any $\ell >n_{m+1}$, we have\[-\varepsilon < x_{n_{m+1}}-a\leqslant x_{\ell}-a\leqslant 0.\]This implies \[|x_l-a|\leqslant \varepsilon.\]Hence $\lim x_n=a$.


Exercise 2.5.3

“If” part. If $-1<b<1$, then $|b|<1$. Then by Example 2.5.3 (it clearly works also for $b=0$) we have $\lim |b|^n=0$. By Algebraic Limit Theorem, we have $\lim -|b|^n=0$.

Note that $$\lim |b|^n=\lim -|b^n|=0$$ and \[-|b|^n\leqslant b^n\leqslant |b|^n,\]by the Squeeze Theorem, we have $\lim b^n=0$.

“Only if” part. Suppose $\lim b^n=0$. By Exercise 2.3.10, $\lim |b^n|=0$. We argue by contradiction. Suppose otherwise $|b|\geqslant 1$, then\begin{equation}\label{eq2.5.5.11}1\leqslant |b|\leqslant |b|^2\leqslant |b|^3\leqslant \cdots.\end{equation} Hence the sequence $|b|^n$ is increasing.

If it is not bounded, then $(|b|^n)$ diverges, contradicting with $\lim |b^n|=0$.

If it is bounded, then it converges by Monotone Convergence Theorem. Moreover, by Order Limit Theorem and \eqref{eq2.5.5.11}, we have $$\lim |b^n|\geqslant 1$$which again contradicts with $\lim |b^n|=0$.

Therefore the assumption $|b|\geqslant 1$ is impossible. Hence $|b|<1$, namely $-1< b < 1$.


Exercise 2.5.5

Suppose $(a_n)$ does not converge to $a$, then there exists $\delta>0$ such that for all $N\in\mathbf N$ there exists $n>N$ satisfying\[|a_n-a|\geqslant \delta.\]We can get a subsequence $(a_{n_k})$ such that \[|a_{n_k}-a|\geqslant \delta\] for all $k\in\mathbf N$ as follows.

Let $N=1$, then we can find $n_1> 1$ such that $|a_{n_1}-a|\geqslant \delta$. Then take $N=n_1$, we can find $n_2> N=n_1$ such that $|a_{n_2}-a|\geqslant \delta$. Then take $N=n_2$, we can find $n_3> N=n_2$ such that $|a_{n_3}-a|\geqslant \delta$. Continuing with this procedure, we get a subsequence $(a_{n_k})$ such that \[|a_{n_k}-a|\geqslant \delta\] for all $k\in\mathbf N$.

Since $(a_n)$ is bounded, therefore there exists a subsequence $(a_{m_k})$ of $(a_{n_k})$ such that $\lim a_{m_k}$ exists and equal to $a$ by assumption. Then by Order Limit Theorem, we have\[0=|\lim a_{m_k}-a|\geqslant \delta.\]Hence we get a contradiction, thus completing the proof.


Exercise 2.5.6

If $b=0$, then $b^{1/n}=0$ for all $n$. Hence $\lim b^{1/n}=0$.

If $b\geqslant 1$, then it is clear\[b\geqslant b^{1/2}\geqslant b^{1/3}\geqslant \cdots\geqslant b^{1/n}\geqslant \cdots \geqslant 1.\]Therefore, $(b^{1/n})$ is decreasing and bounded below. Hence by Monotone Convergence Theorem, $\lim b^{1/n}$ exists. Let $\lim b^{1/n}=x$. By Order Limit Theorem, we also have $x\geqslant 1$.

It follows from Theorem 2.5.2 that  $\lim b^{1/(2n)}=x$ (which is a subsequence).

Since $b^{1/(2n)}b^{1/(2n)}=b^{1/n}$, by Algebraic Limit Theorem, we have\[x^2=x.\]Hence $x=0$ or $1$. But $x\geqslant 1$. Hence $x=1$, namely $\lim b^{1/n}=1$.

If $b\leqslant 1$, then $1/b\geqslant 1$. Hence $\lim (1/b)^{1/n}=1$ by the above. Since $(1/b)^{1/n}=1/b^{1/n}$, by Algebraic Limit Theorem, we have\[\lim b^{1/n}=\frac{1}{\lim (1/b)^{1/n}}=1.\]

Therefore $\lim b^{1/n}=1$ if $b\ne 0$ and $\lim b^{1/n}=0$ if $b=0$.


Exercise 2.5.7

“If” part. If $-1<b<1$, then $|b|<1$. Then by Example 2.5.3 (it clearly works also for $b=0$) we have $\lim |b|^n=0$. By Algebraic Limit Theorem, we have $\lim -|b|^n=0$.

Note that $$\lim |b|^n=\lim -|b^n|=0$$ and \[-|b|^n\leqslant b^n\leqslant |b|^n,\]by the Squeeze Theorem, we have $\lim b^n=0$.

“Only if” part. Suppose $\lim b^n=0$. By Exercise 2.3.10, $\lim |b^n|=0$. We argue by contradiction. Suppose otherwise $|b|\geqslant 1$, then\begin{equation}\label{eq2.5.7.11}1\leqslant |b|\leqslant |b|^2\leqslant |b|^3\leqslant \cdots.\end{equation} Hence the sequence $|b|^n$ is increasing.

If it is not bounded, then $(|b|^n)$ diverges, contradicting with $\lim |b^n|=0$.

If it is bounded, then it converges by Monotone Convergence Theorem. Moreover, by Order Limit Theorem and \eqref{eq2.5.7.11}, we have $$\lim |b^n|\geqslant 1$$which again contradicts with $\lim |b^n|=0$.

Therefore the assumption $|b|\geqslant 1$ is impossible. Hence $|b|<1$, namely $-1< b < 1$.


Exercise 2.5.8

(a) Zero peak term: Take the sequence $(a_n)$ such that $a_n=n$.

One peak term: Take the sequence $(a_n)$ such that\[a_1=2,\quad a_n=1-\frac{1}{n},~n\geqslant 2.\]Two peak terms: Take the sequence $(a_n)$ such that\[a_1=a_2=2,\quad a_n=1-\frac{1}{n},~n\geqslant 3.\]Infinitely many peak terms but not monotone: Take the sequence $(a_n)$ such that\[a_n=(-1)^n\left(1+\frac{1}{n}\right).\]

(b) If the sequence $(x_n)$ has infinitely many peak terms, then take the subsequence consisting of peak terms. It is clear this subsequence is decreasing and bounded by assumption. It follows from the Monotone Convergence Theorem that this subsequence converges.

If the sequence $(x_n)$ has only finitely many peak terms. Let $(x_N)$ be the last peak terms. Take $n_1=N+1$. Since $x_{n_1}$ is not a peak term, there exists $n_2> n_1$ such that $x_{n_2} > x_{n_1}$. Again since $x_{n_2}$ is not a peak term, there exists $n_3> n_2$ such that $x_{n_3} > x_{n_2}$. Continuing this procedure, we find a subsequence $(x_{n_k})$ who is increasing. Since by assumption the subsequence is bounded, it follows from the Monotone Convergence Theorem that this subsequence $(x_{n_k})$ converges.


Exercise 2.5.9

Linearity

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