Assignment 5
Table of contents
url: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/resources/mit18_100af20_hw5/
Chapters and exercises given with a numbering are from Basic Analysis: Introduction to Real Analysis (Vol I) by J. Lebl.
Exercise 2.2.9
Suppose \(\{x_n\}_{n=1}^\infty\) is a sequence, $x \in \mathbb{R}$, and $x_n \ne x$ for all $n \in \mathbb{N}$. Suppose the limit
\[L := \lim_{n \to \infty} \frac{\vert x_{n+1} - x\vert }{\vert x_n - x\vert }\]exists and $L < 1$. Show that ${x_n}_{n=1}^\infty$ converges to $x$.
Since $\lim_{n \to \infty} \frac{\vert x_{n+1} - x\vert }{\vert x_n - x\vert } = L < 1$, let $\varepsilon = 1 - L$, then $\exists M \in \mathbb{N}$, such that for $n > M$, $\frac{\vert x_{n+1} - x\vert }{\vert x_n - x\vert } < L + \varepsilon = 1$. Thus $\vert x_{n+1} - x\vert < \vert x_n - x\vert =: d_n$, so ${d_n}$ is monotone decreasing and bounded below by 0. Therefore it converges, let $\lim_{n \to \infty} \vert x_n - x\vert = D \ge 0$.
If $D > 0$, then by algebraic operations
\[L = \lim_{n \to \infty} \left\vert \frac{x_{n+1} - x}{x_n - x} \right\vert = \frac{\lim_{n \to \infty} \vert x_{n+1} - x\vert }{\lim_{n \to \infty} \vert x_n - x\vert } = \frac{D}{D} = 1\]which contradicts with $L < 1$. Thus $D = 0$, $\lim_{n \to \infty} x_n = x$.
Exercise 2.3.5
(a) Let $x_n := \frac{(-1)^n}{n}$. Find $\limsup_{n \to \infty} x_n$ and $\liminf_{n \to \infty} x_n$.
$\sup{x_k \mid k \ge n} = \frac{1}{n} \to 0 \Rightarrow \limsup_{n \to \infty} x_n = 0$
$\inf{x_k \mid k \ge n} = -\frac{1}{n} \to 0 \Rightarrow \liminf_{n \to \infty} x_n = 0$
(b) Let $x_n := \frac{(n - 1)(-1)^n}{n}$. Find $\limsup_{n \to \infty} x_n$ and $\liminf_{n \to \infty} x_n$.
$x_n = (-1)^n \left(1 - \frac{1}{n}\right)$
\[\sup\{x_k \mid k \ge n\} = \sup\left\{1 - \frac{1}{k} \mid k \ge n\right\} \to 1, \limsup_{n \to \infty} x_n = 1\] \[\inf\{x_k \mid k \ge n\} = \inf\left\{-1 + \frac{1}{k} \mid k \ge n\right\} \to -1, \liminf_{n \to \infty} x_n = -1\]
Exercise 2.3.6
Let \(\{x_n\}_{n=1}^\infty\) and \(\{y_n\}_{n=1}^\infty\) be bounded sequences such that $x_n \le y_n$ for all $n$. Show
\[\limsup_{n \to \infty} x_n \le \limsup_{n \to \infty} y_n \quad \text{and} \quad \liminf_{n \to \infty} x_n \le \liminf_{n \to \infty} y_n.\]$\forall n \in \mathbb{N}, \ x_n \le y_n \Rightarrow \sup{x_k \mid k \ge n} \le \sup{y_k \mid k \ge n} \Rightarrow \limsup_{n \to \infty} x_n \le \limsup_{n \to \infty} y_n$. $\liminf$ is proved similarly.
Exercise 2.3.7
Let
\[\{x_n\}_{n=1}^{\infty}\]and
\[\{y_n\}_{n=1}^{\infty}\]be bounded sequences.
a) Show that \(\{x_n + y_n\}_{n=1}^{\infty}\) is bounded.
$\exists M_x, M_y \in \mathbb{N}$, s.t. ∀$n \in \mathbb{N}$, $\vert x_n\vert \le M_x$, $\vert y_n\vert \le M_y$
Let $M = M_x + M_y + 1$, then $\vert x_n + y_n\vert \le \vert x_n\vert + \vert y_n\vert \le M_x + M_y < M$
b) Show that
\(\liminf_{n \to \infty} x_n + \liminf_{n \to \infty} y_n \le \liminf_{n \to \infty} (x_n + y_n)\)
Hint: One proof is to find a subsequence
\[\{x_{n_m} + y_{n_m}\}_{m=1}^{\infty}\]of
\[\{x_n + y_n\}_{n=1}^{\infty}\]that converges. Then find a subsequence \(\{x_{n_m}\}_{m=1}^{\infty}\), \(\{y_{n_m}\}_{m=1}^{\infty}\) that converges.
Since ${x_n + y_n}$ is bounded, by Bolzano–Weierstrass theorem, exists a subsequence ${x_{n_m} + y_{n_m}}$ with limit
\[\lim_{m \to \infty} (x_{n_m} + y_{n_m}) = \liminf_{n \to \infty} (x_n + y_n)\]${x_{n_m}} \subseteq {x_n}$ is bounded, exists $\lim x_{n_m}$. ${y_{n_{m_i}}} \subseteq {y_{n_m}}$ is bounded, exists $\lim_{i\to\infty} y_{n_{m_i}} = \lim_{m \to \infty} y_{n_m}$. This is to make $x_n$ and $y_n$ share the same index $n_{m_i}$.
\[\lim_{i \to \infty} x_{n_{m_i}} + \lim_{i \to \infty} y_{n_{m_i}} = \liminf_{n \to \infty} (x_n + y_n)\]Since $n_{m_i} \ge m_i \ge i \ge 1$, ${x_k \mid k \ge n_{m_i}} \subseteq {x_k \mid k \ge m_i} \subseteq {x_k \mid k \ge i}$,
\[x_{n_m} \ge \inf \{x_k \mid k \ge n_{m_i}\} \ge \inf \{x_k \mid k \ge m_i\} \ge \inf \{x_k \mid k \ge i\}\]Take limits, $\lim_{i \to \infty} x_{n_{m_i}} \ge \lim_{i \to \infty} \inf {x_k \mid k \ge i} = \liminf_{n \to \infty} x_n$. Similarly, \(\lim_{i \to \infty} y_{n_{m_i}} \ge \liminf_{n \to \infty} y_n\).
So
\(\liminf_{n \to \infty} x_n + \liminf_{n \to \infty} y_n \le \liminf_{n \to \infty} (x_n + y_n)\)
c) Find an explicit ${x_n}$ and ${y_n}$ such that
\(\liminf_{n \to \infty} x_n + \liminf_{n \to \infty} y_n < \liminf_{n \to \infty} (x_n + y_n)\)
Hint: Look for examples that do not have a limit.
Define $x_n = (-1)^n$, $y_n = (-1)^{n+1}$
\[\liminf_{n \to \infty} x_n = \liminf_{n \to \infty} y_n = -1\]But
\(\liminf_{n \to \infty} (x_n + y_n) = 0 \Rightarrow -2 < 0\)
Problem 5
Let ${x_n}$ be a bounded sequence of real numbers. Prove \(\lim_{n \to \infty} x_n = 0,\) if and only if \(\limsup_{n \to \infty} \vert x_n\vert = 0.\)
($\Leftarrow$) Given $\limsup_{n \to \infty} \vert x_n\vert = 0$, then $\forall \varepsilon > 0$, $\exists M \in \mathbb{N}$, such that $\forall n > M$,
\[\left\vert \sup \{ \vert x_k\vert : k \geq n \} \right\vert = \sup \{ \vert x_k\vert : k \geq n \} < \varepsilon\]Since $\vert x_n\vert \leq \sup { \vert x_k\vert : k \geq n }$, $\vert x_n\vert < \varepsilon \Rightarrow \lim_{n \to \infty} x_n = 0$
($\Rightarrow$) Given $\lim_{n \to \infty} x_n = 0$. $\forall \varepsilon > 0$, $\exists N \in \mathbb{N}$, such that for all $n > N$, $\vert x_n\vert < \varepsilon$.
Now consider the set \(\{x_k : k \geq n\}\). Let $\varepsilon > 0$, $\forall k \geq n > N$, $\vert x_k\vert < \varepsilon$.
So for all $n > N$, \(\sup \{ \vert x_k\vert : k \geq n \} < \varepsilon \Rightarrow \limsup_{n \to \infty} \vert x_n\vert = 0\)
Problem 6
Does there exist a sequence \(\{x_n\}\) such that
\[\liminf_{n \to \infty} x_n = -1, \quad \lim_{n \to \infty} x_n = 0, \quad \limsup_{n \to \infty} x_n = 1\ ?\]Either give an example, or explain why no such example exists.
No. \(\liminf_{n \to \infty} x_n = \lim_{n \to \infty} \inf \{ x_k : k \geq n \}\),
where \(\inf \{ x_k : k \geq n \}\) is a subsequence of \(\{x_n\}\).
The limit of a subsequence should equal $\lim_{n \to \infty} x_n$. Thus, $\liminf_{n \to \infty} x_n = \limsup_{n \to \infty} x_n = 0$.
Exercise 2.4.8
True or false, prove or find a counterexample: If \(\{x_n\}_{n=1}^\infty\) is a Cauchy sequence, then there exists an $M$ such that for all $n \geq M$, we have $\vert x_{n+1} - x_n\vert \leq \vert x_n - x_{n-1}\vert$.
$x_n = \frac{1}{\log(n)} \to 0$, so \(\{x_n\}\) is Cauchy. $f(n) := \frac{1}{\log(n)}, f’(n) = -\frac{1}{n (\log n)^2} < 0, f’‘(n) = \frac{2 + \log n}{n^2 (\log n)^3} > 0 \quad (n \geq 1)$. So $f(n)$ is convex in $(1, +\infty)$. Thus $\vert x_{n+1} - x_n\vert > \vert x_n - x_{n-1}\vert $.