# 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

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