Example: If \(a>1\), then \(\lim_{n\to\infty} \dfrac{1}{a^n} = 0\).
\[\dfrac{1}{(1+(a-1))^n} = \dfrac{1}{(1+b)^n} \le \dfrac{1}{1 + nb}\]Ex1 (squeeze theorem): \(\lim_{n\to\infty}a_n = \lim_{n\to\infty}b_n = L\), and \(a_n \le c_n \le b_n\), then \(\lim_{n\to\infty}c_n = L\).
\[\forall \varepsilon > 0, \exists N_a, N_b \in \mathbb{N} [n \ge \max(N_a, N_b) \implies L-\epsilon < a_n \le b_n < L + \epsilon]\] \[\forall \varepsilon > 0, \exists N_a, N_b \in \mathbb{N} [n \ge \max(N_a, N_b) \implies L-\epsilon < a_n \le c_n \le b_n < L + \epsilon]\]Ex: prove for any \(k \in \mathbb{N}\), and real number \(a > 1\), \(\lim_{n\to\infty} \dfrac{n^k}{a^n} = 0\)
Let \(a = 1+b\) where \(b > 0\).
\[\text{ for any } n > k+1, \quad a^n = (1+b)^n \ge \binom{n}{k+1}b^{k+1} \ge \dfrac{b^{k+1}}{(k+1)!} (n-k)^{k+1}\] \[0 \le \dfrac{n^k}{a^n} \le \dfrac{n^k}{(n-k)^k} \dfrac{1}{n-k} \dfrac{k!}{b^k}\]By squeeze theorem, we have \(\lim_{n\to\infty} \dfrac{n^k}{a^n} = 0\).
Monotone
(1) nondecreasingly monotone / increasing: \(\forall n \in N, a_n \le a_{n+1}\)
nonincreasingly monotone / decreasing: \(\forall n \in N, a_n \ge a_{n+1}\)
\[a_n \nearrow \quad a_n \searrow\](2) strictly increasing (resp. strictly decreasing)
\[a_n \nearrow\nearrow \quad a_n \searrow\searrow\]Theorem: (boundedness from above \(+ \nearrow \implies\) convergence)
Since it is boundedness from above, let \(L = \sup \{a_n\}\).
\[\forall \varepsilon> 0, L - \varepsilon \text{ is not upper bound}, \exists N \in \mathbb{N}, [n \ge N \implies a_n \ge a_{N} > L - \varepsilon]\]plus \(a_n \le L\), we have \(\vert a_n - L \vert < \epsilon\).
The Decimal Expression
Examples: (1) A decimal expression gives a real number.
\[0.d_1 d_2 d_3 \dots\] \[a_1 = 0.d_1, a_2 = 0.d_1 d_2, \dots\] \[\forall n \in \mathbb{N} \, [a_n \in \mathbb{Q}]\] \[a_n \nearrow, \text{ it has uppber bound } 1\]one real number may have different decimal expression, 1, 0.999999
The Natural Base
(2) The natural base \(e\)
\[e = \lim_{n\to\infty} (1 + \dfrac{1}{n})^n\] \[\begin{align} a_n &= \binom{n}{0} + \binom{n}{1}\dfrac{1}{n} + \binom{n}{2}\dfrac{1}{n^2} + \dots + \binom{n}{n}\dfrac{1}{n^n}\\ &= \sum_{j=0}^n \binom{n}{j}\dfrac{1}{n^j} \end{align}\] \[\binom{n}{j} = \dfrac{n!}{j!(n-j)!}\] \[\begin{align} &\dfrac{n!}{j!(n-j)!} \dfrac{1}{n^j} = \dfrac{1}{j!} \dfrac{n (n-1) \dots (n-j+1)}{n^j}\\ =& \dfrac{1}{j!} (1-\frac{1}{n})(1-\frac{2}{n})\dots (1 - \dfrac{j-1}{n}) \end{align}\] \[a_n = 1 + 1 + \dfrac{1}{2!}(1-\frac{1}{n}) + \dfrac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n}) + \dots + \dfrac{1}{n!}(1-\frac{1}{n})\dots (1-\frac{n-1}{n})\] \[a_{n+1} = 1 + 1 + \dfrac{1}{2!}(1-\frac{1}{n+1}) + \dfrac{1}{3!}(1-\frac{1}{n+1})(1-\frac{2}{n+1}) + \dots + \dfrac{1}{n!}(1-\frac{1}{n+1})\dots (1-\frac{n-1}{n+1}) + \dfrac{1}{(n+1)!}(1-\frac{1}{n+1})\dots (1-\frac{n}{n+1})\]thus
\[a_{n+1} > a_{n} \quad a_{n} \nearrow\nearrow\]and
\[\begin{align} a_n &< 1 + 1 + \frac{1}{2!} + \dots + \frac{1}{n!}\\ &< 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \dots \frac{1}{2^{n-1}}\\ &= 1 + \dfrac{1-\frac{1}{2^n}}{1-\frac{1}{2}} < 3 \end{align}\]in the near future we will see that
\[e = \lim_{n\to\infty}(1 + 1 + \frac{1}{2!} + \dots + \frac{1}{n!})\]Nested Intervals
Definition: A sequence of intervals \(I_n (n \in \mathbb{N})\) is nested if \(I_n \ne \emptyset\) and \(I_{n+1} \subseteq I_n\) for all \(n \in \mathbb{N}\).
Question: \(\bigcap\limits_{n\in\mathbb{N}} I_n \ne \emptyset\)?
Sometimes it becomes emptyset.
(1) \(I_n = (0, \frac{1}{n})\)
(2) \(I_n = [n, \infty)\)
Theorem (of nested intervals)
If \(I_{n} (n \in \mathbb{N})\) is a sequence of bounded closed nested intervals (\([a_n, b_n]\)), then \(\bigcap\limits_{n\in\mathbb{N}}I_n \ne \emptyset\).
Proof:
Write \(I_n = [a_n, b_n] (n \in \mathbb{N})\), we know \(a_n \le b_n, \quad a_n \nearrow, \quad b_n \searrow\). We know
\[\forall n, m \in \mathbb{N}, a_n \le a_{\max{(n,m)}} \le b_{\max_{(n,m)}} \le b_{m}\]In other words, for every \(m \in \mathbb{N}\), \(b_m\) is a upper bound of \(\{a_n \vert n \in \mathbb{N}\}\). Let \(c=\lim_{n\to\infty}a_n\). Then \(c\) is the least uppber bound, \(c \ge a_n\) for all \(n\). For all \(m\), \(a_n \le b_m\), take the limits \(\lim_{n\to\infty} a_n \le \lim_{n\to\infty} b_m\), thus \(c \le b_m\) for all \(m\).
Ex2: (1) What if \(I_n = (a_n, b_n)\) nested, but \(a_n \nearrow\nearrow\) and \(b_n \searrow\searrow\)?
In this case the intersections are non-empty.
\[\forall n, m \in \mathbb{N}, a_n < a_{\max(n,m)} < b_{\max(n,m)} < b_{m}\]Let \(c = \lim_{n\to\infty} a_n\), then \(c\) is the least uppber bound, \(c \ge a_{n+1} > a_{n}\) for all \(n\). Also \(c \le b_{m+1} < b_m\)
(2) \(I_n = (a_n, \infty)\), nested and \(\{a_n\}\) bounded above.
In this case the intersections are non-empty. Let \(c\) be an upper bound. For all \(n\), \(c \ge a_{n}\), thus \(c + 1 \ge a_{n} + 1 > a_n\)
Equivalence
gapless is equivalent to Weierstrass, to \((\text{ mono } + \text{ Archimedean })\) to \((\text{ nexted interval } + \text{ Archimedean })\).
These expressions are used to determine if we can define something and it is a real number. For example, if we can find a bounded increasing sequence, then we know there is a real number to be its limit.
Cauchy Sequence
A sequence \(a_n (n \in \mathbb{N})\) in \(\mathbb{R}\) is a Cauchy sequence if
\[\forall \varepsilon > 0\, \exists N \in \mathbb{N} \, [n, m \ge N \implies \vert a_n - a_m \vert < \varepsilon]\]It is easy to see
convergent \(\implies\) Cauchy
Cauchy \(\implies\) bounded
Theorem: Cauchy \(\iff\) convergent.
To prove the theorem, we need the definition of upper and lower sequence.
Upper And Lower Sequence
Definition:
For a bounded sequence \(a_n\), let
\[u_n = \sup \{a_m \vert m \ge n\}\] \[l_n = \inf \{a_m \vert m \ge n\}\]we call \(u_n\) the upper sequence of \(a_n\), \(l_n\) the lower sequence of \(a_n\). We want to show for the bounded sequence \(a_n\), the upper sequence and lower sequence converge: \(\lim_{n\to\infty} u_n, \lim_{n\to\infty} l_n\) exists. It is easy to see \(u_n \searrow, \, l_n \nearrow\). Also \(l_n \le a_n \le u_n\).
For example, to show \(u_n \searrow\), we need to show \(\forall i \ge 1, u_{i+1} \le u_{i}\). Were this false, \(\exists i \ge 1, u_{i} < u_{i+1}\).
\[u_{i} < u_{i+1} = \sup \{a_m \vert m \ge i + 1\}\] \[u_{i} \text{ is not upper bound of } \{a_m \vert m \ge i + 1\}\] \[u_{i} \text{ is upper bound of } \{a_m \vert m \ge i\} \implies u_{i} \text{ is upper bound of } \{a_m \vert m \ge i+1\}\]Also, the sequence \(\{u_n\}\), \(\{l_n\}\) is bounded. For example, to show \(\{u_n\}\) is bounded, since \(u_n \searrow\), we only need to show \(\{u_n\}\) has lower bound. Let \(A\) be a lower bound of \(\{a_n\}\), we want to show \(A\) is also a lower bound of \(\{u_n\}\). Were this false, \(A\) is not a lower bound of \(\{u_n\}\)
\[\exists i, u_i < A\] \[\exists i, a_i \le \sup \{a_m \vert m \ge i\} = u_i < A\]Then we have \(a_i < A\), this is impossible, since \(A\) is lower bound of \(\{a_n\}\).
\[\overline{\lim_{n\to\infty}} a_{n}=\limsup_{n\to\infty} a_n := \lim_{n\to\infty} u_n\] \[\underline{\lim}\limits_{n\to\infty} \liminf_{n\to\infty} a_n := \lim_{n\to\infty} l_n\]Ex4: \(a_n\) converges \(\iff\) \(\limsup_{n\to\infty} a_n = \liminf_{n\to\infty} a_n \ne \infty\).
And if it is the case, \(\lim_{n\to\infty}a_n = \limsup_{n\to\infty} a_n = \limsup_{n\to\infty} a_n\)
We have \(l_n \le a_n \le u_n\)
\(\Leftarrow\) is from squeeze theorem.
\(\implies\), Let \(\lim_{n\to\infty}a_n\) = A
\[\forall \epsilon > 0\, \exists N \in\mathbb{N}\, [n \ge N \implies A - \varepsilon < a_n < A + \varepsilon ]\]thus \(A - \varepsilon\) is a lower bound of \(\{a_m \vert m \ge N\}\), and \(l_N\) is the greatest lower bound of \(\{a_m \vert m \ge N\}\), thus \(A - \varepsilon \le l_N \le l_n\) for all \(n \ge N\). Similarly
\[\forall \epsilon > 0\, \exists N \in\mathbb{N}\, [n \ge N \implies A - \varepsilon \le l_{N} \le l_n \le a_n \le u_n \le u_N \le A + \varepsilon ]\]thus \(\limsup_{n\to\infty} a_n = \liminf_{n\to\infty} a_n = A\).