##### 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.

## Michael

30 Jul 2022Small detail, in 2.4.1 part a, I think when you say "By assumption we have $x_n - x_(n+1) > 0$, $4 - x_n > 4 - x_1 > 1$, and $4 - x_(n+1) > 4 - x_1 > 1$" It should rather say "...$x_n - x_(n+1) > 0$, $4 - x_n > 4 - x_1 = 1$, and $4 - x_(n+1) > 4 - x_1 = 1$"

Cheers

## Michael

30 Jul 2022Sorry for my shitty markup! Still learning :)

## LurkingPhysicist

23 Apr 2022Hi there, just wanted to thank you for this amazing site but also point out, as Eduardo said, that it looks like your proof of 2.4.4 uses the AoC even though it asks not to. I tried a proof that basically just combines the same proof of MCT thrm 2.4.2 with example 1.3.3 on page 16. This allows you to essentially prove the Archimdean property for 1/n < x and so that proves (ii), and like in the original proof of thrm 1.4.2 but opposite way, just set x = 1/x in the inequality and you get back (i) of thrm 1.4.2

Thank you so much again though, this has been so helpful for self-study and those of us too poor to afford any kind of high education at the moment.

## Eduardo Freire

17 Apr 2021In exercise 2.4.4 you used the proof of the Monotone Convergence Theorem to justify the fact that $a \geq a_n$, but the proof of MCT requires taking the least upper bound of a set, so it makes use of the Axiom of Completeness, which you are not supposed to do in this exercise. To avoid this in my solution, I used the fact that $a_{n+m} \geq a_n$ for all $n,m \in \mathbb{N}$ together with the Order Limit Theorem to get $\lim_{m \rightarrow \infty} a_{n+m} = \lim_{m \rightarrow \infty} a_m = a \geq \lim_{m \rightarrow \infty}a_n = a_n$ for every $n \in \mathbb{N}$. Do you think this is correct?

Also, thanks a lot for this website, it has been extremely helpful in self-learning analysis.