MATH 447: Real Variables - Homework #2

Proof. $\forall{} s_n, s_n \neq{} 0, \lim_{n \to \infty} | \frac{s_{n+1}}{s_n} | = L.$ $\forall{} \epsilon{} > 0, \exists{} N \in{} \mathbb{N} \text{s.t.} n > N \implies{} | \frac{s_{n+1}}{s_n} - L | < \epsilon{}.$ This implies that $\frac{s_{n+1}}{s_n} < \epsilon{} + L.$ Then we select a value $a$ such that $L < a < 1.$ $$\begin{align} s_{n+1} &< (\epsilon + L) s_n \tag{1} \\ a &= \epsilon + L \tag{2} \end{align}$$ To show that $|s_n| < a^{n-N} |s_N|$, we will show the following: $$\begin{align} |s_n| &= \left| \frac{s_n}{s_{n-1}} \cdot \frac{s_{n-1}}{s_{n-2}} \cdot \dots \cdot \frac{s_{N+1}}{s_N} \cdot |s_N| \right| \tag{4} \\ |s_n| &< a^{n-N} |s_N| \tag{5} \end{align}$$ $\frac{1}{a^N} |s_N|$ is a constant, so we can rename it as the following: $$\begin{align} C = \frac{1}{a^N} |s_N| \tag{6} \end{align}$$ Then, we can say the following: $$\begin{align} |s_n| < a^n \tag{7} \end{align}$$ If $a < 1$, we know that $\lim_{n \to \infty} a^n = 0.$ By Theorem 9.7 (b) in Ross, we know that $\lim_{n \to \infty} a^n = 0.$ Therefore, by Theorem 9.2 in Ross, $\lim_{n \to \infty} |s_n| = 0.$

Proof. Let $t_n = \frac{1}{|s_n|}$. Our goal is to prove $\lim_{n \to \infty} t_n = 0.$ Then: $$\begin{align} L &= \lim_{n \to \infty} \left| \frac{s_{n+1}}{s_n} \right| \notag \\ &= \lim_{n \to \infty} \left| \frac{t_n}{t_{n+1}} \right| \tag{9} \end{align}$$ $\forall{} \epsilon{} > 0, \exists{} N \in{} \mathbb{N} \text{s.t.} n > N \implies{} | \frac{t_n}{t_{n+1}} - L | < \epsilon{}.$ $$\begin{align} \frac{t_n}{t_{n+1}} &< \left( \frac{\epsilon + L}{a} \right) t_n \tag{11} \end{align}$$ Applying the same process as in (a), we can say that $L < a < 1.$ Then: $$\begin{align} t_{n+1} &< at_n \tag{12} \\ \frac{t_{n+1}}{t_n} &< a \tag{13} \\ |t_n| &= \left| \frac{t_n}{t_{n-1}} \cdot \frac{t_{n-1}}{t_{n-2}} \cdot \dots \cdot \frac{t_{N+1}}{t_N} \cdot |t_N| \right| \tag{14} \\ C &= a^{-N} |t_N| \tag{15} \end{align}$$ We can say that $|t_n| < a^n C$ if $a < 1$, so $\lim_{n \to \infty} a^n = 0.$ Then $\lim_{n \to \infty} t_n = 0.$ Therefore: $$\begin{align} \lim_{n \to \infty} \frac{1}{|s_n|} = 0 \tag{16} \end{align}$$ By Theorem 9.10 in Ross, $\lim_{n \to \infty} |s_n| = \infty{}.$

Proof. We begin with $s_n = \lim_{n \to \infty} \frac{a^n}{n^P}, |a| \leq{} 1$. Applying Problem 1: $$\begin{align} \left| \frac{s_{n+1}}{s_n} \right| &= \frac{a^{n+1}}{(n+1)^P} \cdot \frac{n^P}{a^n} \tag{17} \\ &= a \left( \frac{n^P}{(n+1)^P} \right) \tag{18} \\ &= a \tag{19} \quad \text{(since } \lim_{n \to \infty} \frac{n^P}{(n+1)^P} = 1\text{)}. \end{align}$$ The $\lim_{n \to \infty} | \frac{s_{n+1}}{s_n} |$ exists, so Problem 1 applies. According to the result from Problem 1a, if $\lim_{n \to \infty} | \frac{s_{n+1}}{s_n} | < 1$, then $\lim_{n \to \infty} s_n = 0$. When $|a| \leq{} 1$, $\lim_{n \to \infty} \frac{a^n}{n^P} = a$. Therefore, its limit must be 0 when $|a| \leq{} 1$, as desired.

Proof. We begin with $s_n = \lim_{n \to \infty} \frac{a^n}{n^P}, a > 1$. Applying Problem 1: $$\begin{align} \left| \frac{s_{n+1}}{s_n} \right| &= \frac{a^{n+1}}{(n+1)^P} \cdot \frac{n^P}{a^n} \tag{20} \\ &= a \left( \frac{n^P}{(n+1)^P} \right) \tag{21} \\ &= a \tag{22} \quad \text{(since } \lim_{n \to \infty} \frac{n^P}{(n+1)^P} = 1\text{)}. \end{align}$$ The $\lim_{n \to \infty} | \frac{s_{n+1}}{s_n} |$ exists, so Problem 1 applies. According to the result from Problem 1a, if $\lim_{n \to \infty} | \frac{s_{n+1}}{s_n} | > 1$, then $\lim_{n \to \infty} s_n = +\infty{}$. When $a > 1$, $\lim_{n \to \infty} \frac{a^n}{n^P} = a$. Therefore, its limit must be $+\infty{}$ when $a > 1$, as desired.

Proof. We begin with $s_n = \lim_{n \to \infty} \frac{a^n}{n^P}, a < -1$. To show that $\lim_{n \to \infty} \frac{a^n}{n^P} = \text{DNE}$, we will show that there exists more than one limit for the sequence $s_n$. Let $s_{n_1}$ be the subsequence such that $n$ is even. Let $s_{n_2}$ be the subsequence such that $n$ is not even. To show that a limit is divergent, the following must be satisfied: $$\begin{align} \forall M > 0, \exists N \in \mathbb{N} \text{ s.t. } n \geq N \implies \left( \frac{a^{2n}}{(2n)^P} \right) > M \tag{23} \end{align}$$ Choose $N$ such that $N > \frac{P \ln 2 + \ln M}{2 (\ln{} a - P)}$. Then, for all $n > N$, this implies: $$\begin{align} n(2 \ln a - P) &> P \ln 2 + \ln M \tag{24} \\ 2n \ln a - P \ln n &> \ln (2^P M) \tag{25} \\ \ln \left( \frac{a^{2n}}{N^P} \right) &> \ln (2^P M) \tag{26} \\ \frac{a^{2n}}{(2n)^P} &> M \tag{27} \end{align}$$ This implies that the $\lim_{n \to \infty} s_{n_1}$ is divergent. The same reasoning follows with $\lim_{n \to \infty} s_{n_2}$. We then get: $$\begin{align} \lim_{n \to \infty} s_{n_1} &= \lim_{n \to \infty} \frac{a^{2n}}{(2n)^P} \tag{28} \\ \lim_{n \to \infty} s_{n_2} &= \lim_{n \to \infty} \frac{a^{2n+1}}{(2n+1)^P} \tag{29} \end{align}$$ When $a < -1, n \in{} \mathbb{N}$, $a^{2n}$ is always positive and $a^{2n+1}$ is always negative. We then get: $$\begin{align} \lim_{n \to \infty} s_{n_1} &= \lim_{n \to \infty} \frac{a^{2n}}{(2n)^P} = +\infty \tag{31} \\ \lim_{n \to \infty} s_{n_2} &= \lim_{n \to \infty} \frac{a^{2n+1}}{(2n+1)^P} = -\infty \tag{32} \end{align}$$ There are two subsequences of $s_n$ with two distinct limits, so by Theorem 11.8 iii) in Ross, the limit of $s_n$ with $a < -1$ does not exist.

Proof. Let $s_n$ be a sequence such that $\forall{} n \in{} \mathbb{N}, |s_{n+1} - s_n| < 2^{-n}$. To show that a sequence is Cauchy, we must satisfy the following: $$\begin{align} \forall \epsilon > 0, \exists N \in \mathbb{N} \text{ s.t. } n, m > N \implies |s_n - s_m| < \epsilon. \end{align}$$ We will show this by showing that: $$\begin{align} \forall \epsilon, \exists N_0 \in \mathbb{N} \text{ s.t. } n > N_0 \implies |2^{-n} - 0| < \epsilon. \end{align}$$ Then we can bound $s_n < \epsilon{}$ for all $n, m \in{} \mathbb{N}$, thereby showing $s_n$ is Cauchy. We will solve for $N$ in the expression: $$\begin{align} \frac{1}{2^N} &< \epsilon \tag{34} \\ 2^N &> \frac{1}{\epsilon} \tag{35} \\ N &> \log_2 \left( \frac{1}{\epsilon} \right) \tag{36} \end{align}$$ Then, we can choose $N = \log_2{} ( \frac{1}{\epsilon{}} )$. Then, $\forall{} n > N$, $$\begin{align} |s_{n+1} - s_n| &< 2^{-n} \tag{37} \\ |s_m - s_n| &< 2^{-n} < \epsilon \tag{38} \\ |s_m - s_n| &< \epsilon \tag{39} \end{align}$$ Therefore, the sequence $s_n$ is Cauchy, as required.

Proof. We can follow a similar line of reasoning from the previous question. We want $\forall{} n \in{} \mathbb{N}, s_n = |s_{n+1} - s_n| < \frac{1}{n}$. So we will prove that: $$\begin{align} \forall \epsilon > 0, \exists N \text{ s.t. } m, n > N \implies |s_{n+1} - s_n| < \epsilon. \end{align}$$ $$\begin{align} |s_{n+1} - s_n| &< \frac{1}{n} \tag{40} \\ \left| \frac{1}{n} - 0 \right| &< \epsilon \tag{41} \end{align}$$ To determine $N$, we will use algebra as follows: $$\begin{align} \frac{1}{N} &< \epsilon \tag{42} \\ \frac{1}{\epsilon} &< N \tag{43} \end{align}$$ We will choose $N = \frac{1}{\epsilon{}}$. Then, $n > N$ implies: $$\begin{align} \frac{1}{n} &< \epsilon \tag{44} \end{align}$$ And hence, $$\begin{align} \left| \frac{1}{n} - 0 \right| &< \epsilon \tag{45} \\ |s_m - s_n| &< \frac{1}{n} < \epsilon \tag{46} \\ |s_m - s_n| &< \epsilon \tag{47} \end{align}$$ Therefore, the sequence satisfies the Cauchy criterion.

Proof. We will use induction. The base case is as follows: $$\begin{align} s_1 = 1 \tag{53} \end{align}$$ The induction hypothesis is: $$\begin{align} \forall n \geq 1, \frac{1}{2} < s_{n+1} < s_n < 1 \tag{54} \end{align}$$ The inductive step is as follows: $$\begin{align} \frac{1}{3}(s_{n+1} + 1) &< s_{n+1} \tag{55} \\ \frac{s_{n+1}}{3} + \frac{1}{3} &< s_{n+1} \tag{56} \\ \frac{1}{3} &< \frac{2s_{n+1}}{3} \tag{57} \\ \frac{1}{2} &< s_{n+1} \tag{58} \end{align}$$ To finish the proof, we need $\frac{1}{3}(s_{n+1} + 1) > \frac{1}{2}$: $$\begin{align} \frac{s_{n+1}}{3} - \frac{1}{6} &> 0 \tag{59} \\ \frac{1}{6}(2s_{n+1} - 1) &> 0 \tag{60} \\ s_{n+1} &> \frac{1}{2} \tag{61} \end{align}$$

Proof. To prove that $\forall{} n \geq{} 1, s_1 = 1, s_n = \frac{1}{3}(s_n + 1)$ is decreasing, we will show that $\forall{} n \geq{} 1, s_{n+1} \leq{} s_n$: $$\begin{align} \frac{1}{3}(s_{n+1} + 1) &< s_n \tag{62} \\ \frac{s_n}{3} + \frac{1}{3} &< s_n \tag{63} \\ \frac{1}{3} &< \frac{2s_n}{3} \tag{64} \\ \frac{1}{2} &< s_n \tag{65} \end{align}$$ The last line was proved by induction in part (b) of this problem. So $s_n$ is decreasing.

Proof. To show that $\lim_{n \to \infty} s_n$ exists, we can state that because $s_n$ is decreasing, it is monotone. Because $s_n > \frac{1}{2}$ for all $n \in{} \mathbb{N}$, the sequence is also bounded. Therefore, by Theorem 10.2 in Ross, $s_n$ converges and must have a limit. Let $\epsilon{} > 0$, $S = \{s_n : n \in{} \mathbb{N}\}$, and $u = \inf{} s_n$. $u + \epsilon{}$ is not a lower bound of $S$, so $\exists{} N \text{s.t.} s_N < u + \epsilon{}$ for all $n \geq{} N$. Thus, $u \leq{} s_n < u + \epsilon{} \implies{} |s_n - u| < \epsilon{}$. From part (b), we proved $\frac{1}{2} < s_{n+1} < s_n < 1$. So $u = \inf{} s_n = \frac{1}{2}$. Therefore, by Theorem 10.2 from Ross, $$\begin{align} \lim_{n \to \infty} s_n = u = \frac{1}{2}. \end{align}$$

Proof. To show that $t_n$ is decreasing, we will show that $t_{n+1} < t_n$ for all $n \in{} \mathbb{N} \text{s.t.} n \geq{} 1$: $$\begin{align} \left[1 - \frac{1}{(n+1)^2}\right] \cdot t_n &< t_n \tag{66} \\ \left[1 - \frac{1}{(n+1)^2}\right] &< 1 \tag{67} \\ -\frac{1}{(n+1)^2} &< 0 \tag{68} \\ 0 &< \frac{1}{(n+1)^2} \tag{69} \end{align}$$ The last line is always true for $n \in{} \mathbb{N}$, so the $\lim_{n \to \infty} t_n$ exists. We have shown that $t_n$ is monotone, and therefore has a limit.

Proof. We will use induction. For the base case, $n = 1$: $$\begin{align} t_n &= \frac{n + 1}{2n} \tag{70} \\ t_1 &= \frac{2}{2} = 1 \tag{71} \end{align}$$ The inductive hypothesis is as follows: $$\begin{align} t_n = \frac{n + 1}{2n} \quad \text{for } n \geq 1 \tag{72} \end{align}$$ The inductive step: $$\begin{align} t_{n+1} &= \frac{(n+1) + 1}{2(n+1)} \tag{74} \\ &= \frac{n + 2}{2(n+1)} \tag{75} \\ &= \left[ 1 - \frac{1}{(n+1)^2} \right] t_n \tag{76} \\ &= \left[ 1 - \frac{1}{(n+1)^2} \right] \left( \frac{n+1}{2n} \right) \tag{77} \\ &= \frac{(n+1)^2 (n+1)}{2n(n+1)^2} - \frac{n+1}{2n(n+1)^2} \tag{78} \\ &= \frac{n+2}{2(n+1)} \tag{79} \end{align}$$ Thus, the proof is complete.

Proof. $$\begin{align} \lim_{n \to \infty} \frac{n + 1}{2n} &= \lim_{n \to \infty} \frac{1 + 1/n}{2} \tag{80} \\ &= \frac{1}{2} \tag{81} \end{align}$$ Equation (80) is established from Theorem 9.10 in Ross.