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Solution to Understanding Analysis Exercise 2.4


Exercise 2.4.1

(a) We show by induction that $(x_n)$ is decreasing by proving $x_n > x_{n+1}$. Note that $x_1=3$ and $x_2=\dfrac{1}{4-3}=1$. Hence $x_1>x_2$. Suppose now that\[3=x_1> x_2>\cdots >x_n> x_{n+1},\]we show that $x_{n+1}> x_{n+2}$. By assumption we have $x_{n}-x_{n+1}> 0$, $4-x_n>4-x_1>1$, and $4-x_{n+1}>4-x_1>1$. Therefore,\begin{align*}x_{n+1}-x_{n+2}=&~\frac{1}{4-x_n}-\frac{1}{4-x_{n+1}}\\ =& ~\frac{(4-x_{n+1})-(4-x_n)}{(4-x_n)(4-x_{n+1})}\\=&~\frac{x_n-x_{n+1}}{(4-x_n)(4-x_{n+1})}>0.\end{align*}Hence, by induction $x_n> x_{n+1}$ holds for all $n\in\mathbf N$, namely $(x_n)$ is decreasing. Since $x_n< 4$ fro all $n\in\mathbf N$, we have $x_{n+1}=\dfrac{1}{4-x_n}>0$. Hence the sequence is bounded. By Monotone Convergence Theorem, the sequence $(x_n)$ converges.

(b) You can prove it by definition of limit. Limit exists means that the sequences is eventually in any $\varepsilon$-neighborhood of limit. Shifting the indices does not change much. New sequence would be still eventually in any $\varepsilon$-neighborhood of limit (the same limit as the origin sequence).

(c) Taking the limit for both sides, we get by Algebraic Limit Theorem that\[\lim x_{n}=\lim x_{n+1}=\frac{1}{4-\lim x_{n}}.\]Solving it we get that $\lim x_n=2\pm \sqrt{3}$. But by Order Limit Theorem we know the limit should be less than $3$ since it is decreasing. Hence $\lim x_n=2-\sqrt{3}$.


Exercise 2.4.2

(a) The limit may not exist since this sequence is not monotone\[y_{n+1}-y_n=-(-y_n-y_{n-1}).\]Or you can see this sequence is alternating (1,2,1,2,1,2,……).

(b)Yes. We first show that this sequence is increasing. We proceed by induction on to prove that $y_{n+1}-y_n>0$. Note that $y_1=1$, $y_2=2$, this is clear for $n=1$. Suppose now $y_{n+1}-y_n>0$ for all $n=1,\cdots,k$. We would like to show that $y_{k+2}-y_{k+1}>0$. It is clear that $y_{k}>0$ and $y_{k+1}>0$. Then \begin{align*}y_{k+2}-y_{k+1}=&~\left(3-\frac{1}{y_{k+1}}\right)-\left(3-\frac{1}{y_{k}}\right)\\=&~ \frac{1}{y_k}-\frac{1}{y_{k+1}}=\frac{y_{k+1}-y_k}{y_ky_{k+1}}>0\end{align*}Hence by induction the sequence $(y_n)$ is increasing. Since $y_n>0$ we get that $y_{n+1}<3$ for all $n\in \mathbf N$. Hence $(y_n)$ is also bounded. By Monotone Convergence Theorem, $(y_n)$ converges.
Now we take the limit for both sides, as for Excercise 2.4.1, we get\[\lim y_n=3-\frac{1}{\lim y_n}.\]Hence $\lim y_n=\dfrac{3\pm \sqrt{5}}{2}$. Because $y_n\geqslant 1$, we see that $\lim y_n \geqslant 1$ by Order Limit Theorem and hence $\lim y_n=\dfrac{3+\sqrt{5}}{2}$.


Exercise 2.4.3

(a) Let $(x_n)$ be the sequence, then we have\[x_{n+1}=\sqrt{2+x_n}.\]We show by induction that $x_n <2$. It is clear that $x_1 =\sqrt{2}< 2$.Suppose $x_n<2$, then\[x_{n+1}=\sqrt{2+x_n}< \sqrt{2+2}=2.\]Hence by induction, we have $x_n <2$ for all $n\in\mathbf N$. We then show that $(x_n)$ is increasing. Note that\begin{align*}x_{n+1}-x_{n}=&~\sqrt{2+x_n}-x_n\\=&~\frac{2+x_n-x_n^2}{\sqrt{2+x_n}+x_n}\\=&~\frac{(2-x_n)(1+x_n)}{\sqrt{2+x_n}+x_n}>0\end{align*}since $0 < x_n < 2$. Therefore $(x_n)$ is increasing. Since $0 < x_n < 2$, this sequence is also bounded. Therefore by Monotone Convergence Theorem $\lim x_n$ exists. Taking the limit on both sides of $x_{n+1}=\sqrt{2+x_n}$, by Exercise 2.3.1, we have\[\lim x_n=\sqrt{2+\lim x_n}.\]Hence $\lim x_n=2$.

(b) Let $(x_n)$ be the sequence, then we have\[x_{n+1}=\sqrt{2x_n}.\]We show by induction that $x_n <2$. It is clear that $x_1 =\sqrt{2}< 2$.Suppose $x_n<2$, then\[x_{n+1}=\sqrt{2x_n}< \sqrt{2\cdot 2}=2.\]Hence by induction, we have $x_n <2$ for all $n\in\mathbf N$. We then show that $(x_n)$ is increasing. Note that\begin{align*}x_{n+1}-x_{n}=&~\sqrt{2x_n}-x_n\\=&~\frac{2x_n-x_n^2}{\sqrt{2x_n}+x_n}\\=&~\frac{(2-x_n)x_n}{\sqrt{2+x_n}+x_n}>0\end{align*}since $0 < x_n < 2$. Therefore $(x_n)$ is increasing. Since $0 < x_n < 2$, this sequence is also bounded. Therefore by Monotone Convergence Theorem $\lim x_n$ exists. Taking the limit on both sides of $x_{n+1}=\sqrt{2 x_n}$, by Exercise 2.3.1, we have\[\lim x_n=\sqrt{2 \lim x_n}.\]Hence $\lim x_n=2$.


Exercise 2.4.4

(a) By the same reason as in the proof of Theorem 1.4.2, we only need to show part (i). We prove it by contradiction. Suppose that there exists a number $x\in\mathbf R$ such that $n\leqslant x$ for all $n\in\mathbf N$. We consider the sequence $(1,2,3,\cdots,n,\cdots)$. It is clear that this sequence is increasing. Since $n\leqslant x$ for all $n\in\mathbf N$, the sequence is also bounded. Therefore by Monotone Convergence Theorem, we know that this sequence is convergent. Let $a$ be the limit.

Since $\lim n=a$. There exists $N\in\mathbf N$ such that $|n-a|<0.5$ for all $n> N$. In particular, we have $|N+1-a|<0.5$ and $|N+2-a|<0.5$. Therefore we have \[|N+1-a|+|N+2-a|<0.5+0.5=1.\]On the other hand, by the Triangle inequality, we have \[|N+1-a|+|N+2-a|\geqslant |N+2-a-(N+1-a)|=1.\]Hence we get a contradiction. Thus part (i) is proved.

(b) We remark that we are allowed to use the Order Limit Theorem (otherwise you can reproduce it by proof of contradiction). Since $(a_n)$ is increasing and bounded, therefore the limit $\lim a_n$ exists. We denote it by $a$. By the proof of Monotone Convergence Theorem, we have $a\geqslant a_n$ for all $n\in\mathbf N$. Moreover, given arbitrary $m\in\mathbf N$, we have $a_n\leqslant b_m$ for all $n\in\mathbf N$. Again by the Order Limit Theorem, $a\leqslant b_m$ for all $m\in \mathbf N$. Therefore, we have $a_n\leqslant a\leqslant b_n$ for all $n\in\mathbf N$. Hence $a\in I_n$ for all $n\in \mathbf N$ and $a\in\cap_{n=1}^\infty I_n$. In particular, $\cap_{n=1}^\infty I_n\ne \emptyset$.


Exercise 2.4.5

(a) Note that\begin{align*}x_{n+1}^2=&~\frac{1}{4}\left(x_n+\frac{2}{x_n}\right)^2\\=&~\frac{1}{4}\left(x_n-\frac{2}{x_n}\right)^2+2\\ \geqslant &~2,\end{align*}hence the first part is proved. Now we consider $x_n-x_{n+1}$, it is clear that $x_n > 0$ for all $n\in \mathbf N$. Hence $x_n\geqslant \sqrt{2}$. We have\begin{align*}x_n-x_{n+1}=&~x_n-\frac{1}{2}\left(x_n+\frac{2}{x_n}\right)\\=&~\frac{1}{2}\left(x_n-\frac{2}{x_n}\right)\\ =&~\frac{x_n^2-2}{2x_n}\geqslant 0.\end{align*}Hence the sequence is decreasing. Since the sequence is also bounded (between 2 and $\sqrt{2}$), by Monotone Convergence Theorem, we have that $\lim x_n$ exists. Taking the limit on both sides of $x_{n+1}=\dfrac{1}{2}\left(x_n+\dfrac{2}{x_n}\right)$, we have\[\lim x_n=\frac{1}{2}\left(\lim x_n+\frac{2}{\lim x_n}\right).\]Solving it we get that $\lim x_n=\pm \sqrt{2}$. Since $x_n\geqslant \sqrt{2}$, by Order Limit Theorem, $\lim x_n\geqslant \sqrt{2}$. Thus $\lim x_n=\sqrt{2}$.

(b) Change the sequence so that $x_1=\sqrt{c}+1$ (any number great than $\sqrt{c}$) and \[x_{n+1}=\frac{1}{2}\left(x_n+\frac{c}{x_n}\right).\]


Exercise 2.4.6

(a) Note that\begin{align*}(x+y)^2-4xy=(x-y)^2\geqslant 0,\end{align*} we have $(x+y)^2\geqslant 4xy$. Hence we get, by taking square root, $x+y\geqslant 2\sqrt{xy}$. In other words, $\dfrac{x+y}{2}\geqslant \sqrt{xy}$.

(b) We show by induction that $0\leqslant x_n\leqslant x_{n+1}\leqslant y_{n+1}\leqslant y_n$. Note that by (a) $$x_2=\sqrt{x_1y_1}\leqslant \frac{x_1+y_1}{2}=y_2.$$ Since $x_1\leqslant y_1$, we also have $$x_1\leqslant \sqrt{x_1y_1}=x_2$$and $$y_2= \frac{x_1+y_1}{2}\leqslant y_1.$$ We then suppose that $0\leqslant x_n\leqslant x_{n+1}\leqslant y_{n+1}\leqslant y_n$. By repeating the case for $n=1$, we can show $0\leqslant x_{n+1}\leqslant x_{n+2}\leqslant y_{n+2}\leqslant y_{n+1}$. Hence the sequence $(x_n)$ is increasing while $(y_n)$ is decreasing. Moreover they are also bounded. Hence, by Monotone Convergence Theorem, $\lim x_n$ and $\lim y_n$ exist and are nonnegative. Taking limit on both sides of $y_{n+1}=\frac{x_n+y_n}{2}$, we get\[\lim y_n=\frac{\lim x_n+\lim y_n}{2}.\]Therefore $\lim x_n=\lim y_n$.


Exercise 2.4.7

(a) By definition, we have $y_n\geqslant a_k$ for all $k\geqslant n$. In particular, $y_n\geqslant a_k$ for all $k\geqslant n+1$. Therefore $y_n$ is an upper bound of $\{a_k:k\geqslant n+1\}$. By the definition of sup, we have $y_n\geqslant y_{n+1}$. Hence sequence $(y_n)$ is decreasing. Since $(a_n)$ is bounded, there exists $M\in\mathbf R$ such that $a_n\geqslant M$ for all $n\in\mathbf N$. Since $y_n\geqslant a_n$, we have $y_n\geqslant M$ for all $n\in\mathbf N$. Therefore $(y_n)$ is bounded. By Monotone Convergence Theorem, the sequence $(y_n)$ converges.

(b) Let $z_n=\inf \{a_k:k\geqslant n\}$. Similar to part (a), this sequence is increasing and bounded. Hence by Monotone Convergence Theorem, the sequence $(z_n)$ converges. Define limit inferior as follows:\[\lim\inf a_n=\lim z_n.\]

(c) Since \[z_n\leqslant a_n \leqslant y_n\]for all $n\in\mathbf N$, by Order Limit Theorem, we have \[\lim \inf a_n=\lim z_n\leqslant \lim y_n=\lim\sup a_n.\]Let $a_n=(-1)^n$, then it is easy to see that $$\lim\sup a_n=1,~~\lim\inf a_n=-1.$$Hence the inequality is strict.

(d) “Only if” part. Recall in part (c) we have \[z_n\leqslant a_n \leqslant y_n\]for all $n\in\mathbf N$. Moreover \[\lim z_n=\lim\inf a_n=\lim\sup a_n=\lim y_n.\]By Squeeze Theorem, we see that $\lim a_n$ exists as well. Moreover, $$\lim a_n=\lim\inf a_n=\lim\sup a_n.$$

“If” part. Let $\varepsilon>0$ be arbitrary. Let $\lim a_n=a$. Then there exists $N\in\mathbf N$ such that for all $n>N$ we have $$a-\varepsilon < a_n < a+\varepsilon.$$ Now taking $n>N$, for all $k>n$, we also have $a_k < a+\varepsilon$. Hence $a+\varepsilon$ is an upper bound of $\{a_k:k>n\}$. Thus $a+\varepsilon \geqslant y_n$. Note that we also have $$y_n\geqslant a_n >a-\varepsilon.$$Therefore\[a+\varepsilon \geqslant y_n > a-\varepsilon \Longrightarrow |y_n-a|\leqslant \varepsilon\]for all $n>N$. Hence $\lim y_n=a$, namely $\lim\sup a_n=\lim a_n$. Similarly, taking $n>N$, for all $k>n$, we have $a_k >a-\varepsilon$. Hence $a-\varepsilon$ is a lower bound of $\{a_k:k>n\}$. Thus $a-\varepsilon \leqslant z_n$. Note that we also have $$z_n\leqslant a_n <a+\varepsilon.$$Therefore\[a-\varepsilon \leqslant z_n <a+\varepsilon \Longrightarrow |z_n-a|\leqslant \varepsilon\]for all $n>N$. Hence $\lim z_n=a$, namely $\lim\inf a_n=\lim a_n$.


Exercise 2.4.8

(a) The partial sum is given by\begin{align*}s_m=&~\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^m}\\=&~\frac{\frac{1}{2}-\frac{1}{2^{m+1}}}{1-\frac{1}{2}}\\=&~1-\frac{1}{2^m}.\end{align*}Therefore the partial sum is less than 1. Thus the sequence is bounded and, by Monotone Convergence Theorem, convergent. Moreover the limit is $$s=\lim_{m\to \infty}\left(1-\frac{1}{2^m}\right)=1.$$

(b) Note that $\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$, the partial sum is given by\begin{align*}s_m=&\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{m(m+1)}\\=&~\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\ =&~ 1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\cdots+\left(-\frac{1}{n}+\frac{1}{n}\right)-\frac{1}{n+1}\\=&~1-\frac{1}{n+1}.\end{align*}Therefore the partial sum is less than 1. Thus the sequence is bounded and, by Monotone Convergence Theorem, convergent. Moreover the limit is $$s=\lim_{m\to \infty}\left(1-\frac{1}{m+1}\right)=1.$$

(c) Note that by the quotient rule of logarithms, we have $\log\left(\dfrac{n+1}{n}\right)=\log(n+1)-\log n$. The partial sum is given by\begin{align*}s_m=&\log\frac{2}{1}+\log\frac{3}{2}+\cdots+\log\frac{n+1}{n}\\=&~(\log 2-\log 1)+(\log 3-\log 2)+\cdots+(\log (n+1)-\log n)\\=&-\log 1+(\log 2-\log 2)+\cdots+(\log n-\log n)+\log(n+1)\\=&\log(n+1)-\log 1=\log(n+1).\end{align*}Therefore the partial sum is not bounded. Hence the sequence is divergent.


Exercise 2.4.9

Let $s_m$ be the partial sum of $\sum_{n=1}^\infty b_n$. Let $\tilde s_m$ be the partial sum of $\sum_{n=0}^\infty 2^nb_{2^n}$. Since $\sum_{n=1}^\infty 2^nb_{2^n}$ diverges, the sequence $(\tilde s_n)$ is not bounded. To show $\sum_{n=1}^\infty b_n$ diverges, it suffices to show that $(s_n)$ is not bounded. We show that $2s_{2^m}>\tilde{s}_m$. Note that $(b_n)$ is decreasing. We have\begin{align*}2s_{2^m}=&~2b_1+2b_2+2b_3+2b_4+\cdots+2b_{2^{m-1}+1}+2b_{2^{m-1}+2}+\cdots+2b_{2^m}\\ =&~ 2b_1+2b_2+(2b_3+2b_4)+\cdots+(2b_{2^{m-1}+1}+2b_{2^{m-1}+2}+\cdots+2b_{2^m})\\ \geqslant &~ 2b_1+2b_2+(2b_4+2b_4)+\cdots+(2b_{2^{m}}+2b_{2^{m}}+\cdots+2b_{2^m})\\=&~2b_1+2b_2+4b_4+8b_8+\cdots+2^mb_{2^m}\\ \geqslant &~b_1+2b_2+2^2b_{2^2}+2^3b_{2^3}+\cdots+2^mb_{2^m}=\tilde{s}_m.\end{align*}Since $(\tilde{s}_n)$ is not bounded (from above), we know that $(s_n)$ is also not bounded (from above). Therefore $(s_n)$ diverges, namely $\sum_{n=1}^\infty b_n$ diverges.


Exercise 2.4.10

(a) We compute the partial product\begin{align*}p_m=&\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\cdots\left(1+\frac{1}{m}\right)\\=&\frac{2}{1}\cdot\frac{3}{2}\cdot\cdots\cdot \frac{m+1}{m}\\=&~\frac{1}{1}\cdot\frac{2}{2}\cdot\cdots\cdot \frac{m}{m}\cdot(m+1)=m+1.\end{align*} Hence this sequence is divergent.

If $a_n=\dfrac{1}{n^2}$, then\[p_1=2,~p_2=\frac{5}{2},~p_3=\frac{25}{9}.\]It should be convergent, see below.

(b) “If” part. We shall use the inequality $1+x\leqslant 3^x$ for all nonnegative $x$. Then we have\begin{align*}p_m=&~(1+a_1)(1+a_2)\cdots(1+a_m)\\ \leqslant &~ 3^{a_1}\cdot 3^{a_2}\cdot \cdots 3^{a_m}\\ =&~ 3^{a_1+\cdots +a_m}.\end{align*}Since $\sum_{n=1}^\infty a_n$ converges, the partial sum $s_m=\sum_{n=1}^m a_m$ is bounded. Therefore, the partial product $p_m$ is also bounded. It is clear that $(p_n)$ is increasing. Therefore by Monotone Convergence Theorem, $(p_n)$ converges.

“Only if” part. Suppose the partial product is convergent, then $(p_n)$ is bounded. Note that\begin{align*}p_m=&(1+a_1)(1+a_2)\cdots(1+a_m)\\ \geqslant &~ 1+a_1+a_2+\cdots+a_m\\=& 1+s_m.\end{align*}Therefore the partial sum $(s_n)$ is also bounded. It is clear that $(s_n)$ is increasing. Therefore by Monotone Convergence Theorem, $(s_n)$ converges. Namely, $\sum_{n=1}^{\infty}a_n$ converges.

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