Some comments on our usage of logic terms
Standard logic terms
\[\forall \varepsilon \in \mathbb{R} \, [\varepsilon > 0 \implies \exists N \in \mathbb{N} \, \forall n \in \mathbb{N} \, [ n \ge N \implies \vert a_n - L \vert < \varepsilon]]\]informal way
\[\forall \varepsilon > 0 \, \exists N \in \mathbb{N} \, [n \ge N \implies \vert a_n - L \vert < \varepsilon]\]Ex. Let \(a_n, b_n (n \in \mathbb{N})\) be two bounded sequence. Then
\[\overline{\lim_{n\to\infty}} (a_n + b_n) \le \overline{\lim_{n\to\infty}} a_n + \overline{\lim_{n\to\infty}}b_n\] \[\underline{\lim}\limits_{n\to\infty} (a_n + b_n) \ge \underline{\lim}\limits_{n\to\infty} a_n + \underline{\lim}\limits_{n\to\infty} b_n\]Let \(u_n = \sup \{a_m \vert m \ge n\}, v_{n} = \sup \{b_m \vert m \ge n\}, y_{n} = \sup \{a_m + b_m \vert m \ge n\}\). Since \(u_n\) is a upper bound of \(\{a_m \vert m \ge n\}\), \(v_n\) is a upper bound of \(\{b_m \vert m \ge n\}\). We have \(\forall m \ge n, u_n \ge a_m, v_n \ge b_m\). Thus
\[\forall m \ge n, u_n + v_n \ge a_m + b_m\]Thus \(u_n + v_n\) is a upper bound of \(\{a_m + b_m \vert m \ge n\}\). Since \(y_n\) is the least upper bound
\[y_n \le u_n + v_n\] \[\lim_{n\to\infty} y_n \le \lim_{n\to\infty} u_n + \lim_{n\to\infty} v_n\]Ex. What about \(\overline{\lim_{n\to\infty}} (a_n \times b_n)\)? \(\overline{\lim_{n\to\infty}} (a_n \div b_n)\)? If the sequences are all positive, we can have the same conclusion.
Cauchy Sequence Convergent
Since Cauchy sequence is bounded, to prove a bounded sequence convergent, we only need to show \(\lim_{n\to\infty} u_n = \lim_{n\to\infty} l_n\), thus we only need to show \(\lim_{n\to\infty} (u_n - l_n) = 0\). For Cauchy sequence
\[\forall \varepsilon > 0, \exists N \in \mathbb{N}, [n \ge N \implies a_N - \varepsilon < a_n < a_{N} + \varepsilon]\] \[\forall \varepsilon > 0, \exists N \in \mathbb{N}, [n \ge N \implies a_N - \varepsilon \le l_N \le l_n \le a_n \le u_n \le u_N \le a_{N} + \varepsilon]\] \[\forall \varepsilon > 0, \exists N \in \mathbb{N}, [n \ge N \implies \vert u_n - l_n - 0 \vert \le 2 \varepsilon]\]thus we finish the proof.
Ex 1. Let \(S \subseteq \mathbb{R}\). Prove that if \(\vert s - s' \vert \le 3\) for all \(s, s'\), then
(1) \(S\) is bounded.
(2) \(\sup S - \inf S \le 3\).
What if \(\vert s - s' \vert < 3\)? We still only have \(\sup S - inf S \le 3\). Consider the example of \(S = (0,3)\).
(1) is obvious. (2) Were this false, \(\sup S - \inf S = a > 3\). thus \(\sup S = \inf S + a\), thus let \(b = \inf S + a - (a-3)/2 < \sup S\), \(c = \inf S + (a-3)/2 > \inf S\). We know \(c < b\) and \(b\) is not upper bound of \(S\), \(c\) is not lower bound of \(S\).
\[\exists x, y \in S, [y < c < b < x]\] \[\vert x - y \vert > b - c = 3\]Series
Question: What does \(1 + \frac{1}{2} + \frac{1}{2^2} + \dots = 2\) mean?
Def: Let \(a_n (n \in \mathbb{N})\) be a sequence of \(\mathbb{R}\). We say that the series \(\sum_{n=1}^{\infty} a_n\) converges to a real number \(S\), if
\[\lim_{n\to\infty} s_n = S, \text{ where } s_n := \sum_{j=1}^{n} a_j\]we usually call \(s_n\) the \(n\)=th partial sum of the series.
What if it is not easy to calculate the closed form of \(s_n\)? Take the examples
\[1 + \frac{1}{2} + \frac{1}{3} + \dots\] \[1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots\]Notation: For a series \(\sum_{n} a_n\) and \(l, m \in \mathbb{N}, l < m\), we call
\[s_{l,m} := \sum_{j=l}^{m} a_j\]the \((l,m)\)-tail.
- \(\sum_{n} a_n\) converges \(\implies \lim_{n\to\infty} a_n = 0\).
We know \(s_n\) is Cauchy sequence
\[\forall \varepsilon > 0, \exists N \in \mathbb{N}, [m,n >= N \implies \vert s_m - s_n \vert < \varepsilon]\] \[\forall \varepsilon > 0, \exists N \in \mathbb{N}, [n \ge N \implies \vert s_{n+1} - s_n \vert = \vert a_{n+1} \vert < \varepsilon]\]Or
\[\lim_{n\to\infty} s_n = \lim_{n\to\infty} s_{n-1} = S\] \[\lim_{n\to\infty} (s_n - s_{n-1}) = \lim_{n\to\infty} a_n = 0\]