[Calculus] 2. Sequences

Proof Proof 1. Suppose that \(L\ne M\), and \(|L-M|=r\) where \(r\) is a positive real number. And let \(\epsilon=\frac{r}{3}=\frac{|L-M|}{3}\).

By definition, there exist \(N_1,N_2\in\N\), such that \(|a_n - L| < \epsilon\) for all \(n\ge N_1\) and \(|a_m - M| < \epsilon\) for all \(m\ge N_2\). Let \(N = \max\{N_1, N_2\}\), then \(n\ge N \Rarr |a_n - L| < \epsilon \land |a_n - M| < \epsilon\), which implies \(|L-M| < 2\epsilon=\frac{|L-M|}{3}\), a contradiction.

Therefore, \(L=M\).

Proof 2. For any given \(\epsilon > 0\), there exists an \(N\in \N\) such that \(n\ge N \Rarr |a_n - L| < \epsilon\) with the hypothesis \(\lim_{n\to\infty} a_n = L\).

We can find out the maximal and minimal of sequence \(a_1, a_2, \cdots, a_{N-1}\), since the sequence is finite, and write them as \(M_{max}, M_{min}\) respectively. And hence \(M_{min} \le a_n \le M_{max}\) for all \(n < N\).

Similarly, let \(U = \max\{M_{max}, L + \epsilon\}, V = \min\{M_{min}, L-\epsilon\}\), then we have: \(V \le a_n \le U\) for all \(n\in\N\), i.e., \({a_n}\) is bounded.

Proof 3. Like proof 2, \(\forall \epsilon >0 \exists N\in\N(n\ge N\Rarr |a_n - L| < \epsilon \land |b_n - M| < \epsilon)\).

\(a_n - b_n \le 0 \Rarr L - M - 2\epsilon < 0\).

Suppose that \(L > M\), and let \(\epsilon = \frac{L - M}{3} > 0\). Then \(L - M - \frac{2}{3}(L - M) < 0 \Rarr L - M < 0\), a contradiction.

Therefore, \(L \le M\).

If \(\le\) is replaced \(<\), the proposition is false. Here is the example. Let \(a_n = \frac{1}{n^2}, b_n = \frac{1}{n}\), it is easy to prove that these two sequences both converging to \(0\) as \(n\to\infty\). □

Proof Proof 1. \(|(a_n\pm b_n) - (L\pm M)| = |(a_n - L) \pm (b_n - M)| \le |a_n - L| + |b_n - M| < 2\epsilon\). (There are some skips)

Proof 2. \(|a_nb_n - LM| = |a_nb_n - Lb_n + Lb_n -LM| \le |a_n - L||b_n| + |L||b_n - M|\)

Choose a real number \(c>0\), such that \(|b_n| \le c \land |L| \le c\). It is guaranteed by 'convergent implies bounded'.

Then, \(|a_nb_n - LM| = \le |a_n - L||b_n| + |L||b_n - M| \le C(|a_n - L| + |b_n - M|) \le 2C\epsilon\).

Proof 3.

Firstly, let \(\epsilon = \frac{|M|}{2}\). Then \(|M| - |b_n| \le |b_n - M| < \epsilon\), i.e., \(|b_n| > \frac{|M|}{2}\).

\begin{align*} \left|\frac{a_n}{b_n} - \frac{L}{M}\right| &= \frac{|M a_n - L b_n|}{|b_n M|}\\ &=\frac{|M a_n - ML + ML - L b_n|}{|b_n M|}\\ &=\frac{|M (a_n - L) + L (M - b_n)}{|b_n M|}\\ &\le \frac{2}{|M|^2}(|M||a_n - L| + |L||M - b_n|)\\ &\le \frac{2\epsilon}{|M|^2}(|M| + |L|) = C\epsilon \end{ali□

Proof \(\forall \epsilon > 0 \exists N_1, N_2 \in\N [n \ge N_1 \Rarr |a_n - L| < \epsilon \land m \ge N_2 \Rarr |b_m - L| <\epsilon]\).

Let \(N = \max\{N_1, N_2\}\). Then \(n \ge N \Rarr L - \epsilon < a_n \le c_n \le b_n < L +\epsilon \Lrarr |c_n - L| < \epsilon\). □

Proof \(\{a_n|n\in\N\}\) has an upper bound, then \(L:=\sup\{a_n|n\in\N\}\) exists.

We claim that \(\lim_{n\to\infty} a_n = L\).

\(\forall \epsilon > 0\), \(L-\epsilon < L\), i.e., \(L-\epsilon\) is not an upper bound, and hence \(\exists N\in\N [L-\epsilon < a_N]\).

\(\forall n\ge N[a_N\le a_n]\) since \(a_n\) is nondecreasingly monotone.

We obtain: \(L-\epsilon < a_N \le a_n \le L < L-\epsilon \Rarr |a_n - L| < \epsilon\). □

Proof Write \(I_n = [a_n, b_n] (n\in\N)\). Then \(I_n\) is nested iff \(a_n \le b_n\), \(a_n\) is nondecreasing and \(b_n\) is nonincreasing.

\(\forall n,m\in\N(n < m)[a_n \le a_{\max\{n,m\}} \le b_{\min\{n,m\}} \le b_m]\). In other words, for every \(m\in\N\), \(b_m\) is an upper bound of \(\{a_n|n\in\N\}\).

Let \(c=\lim_{n\to\infty} a_n\). Then \(c\le b_m\) for all \(m\in\N\).

Similarly, we have \(a_n\le c\) for all \(n\in\N\).

Thus, \(\forall n\in\N[a_n \le c \le b_n]\), i.e., \(c\in\bigcap_{n\in\N}I_n\). □

Proof Proof 1. Same as the theorem's proof.

Proof 2. It is easy to prove that \(\forall a\in \N[a_n \le a_{n+1}]\), which means \(a_n\) is increasing. And hence, \(c := \lim_{n\to\infty} a_n = \sup\{a_n|n\in\N\}\) exists since \(a_n\) is bounded.

Therefore, \(\forall n\in\N[a_n\le c]\). Consider \(c+1\), \(\forall n\in\N[a_n < c < \infty \Rarr c \in I_n]\). □

Proof Proof 1. Let a sequence \(a_n\) be convergent. Then \(\forall\epsilon > 0 \exists N\in\N[n,m\ge N \Rarr |a_n - L| < \epsilon \land |a_m - L| < \epsilon]\). Therefore, \(|a_n - a_m| < 2\epsilon\).

Proof 2. \(\forall \epsilon > 0 \exists N\in\N[n\ge N, m=N \Rarr |a_n - a_N| < \epsilon]\), since \(a_n\) is Cauchy.

Then \(|a_n| = |(a_n - a_N) + a_N| \le |a_n - a_N| + |a_N| < |a_N| + \epsilon\), i.e., for any fixed \(\epsilon\), \(a_n\) is bounded by \(|a_N| + \epsilon\) for every \(n\ge N\). Let M = \(\max\{|a_1|, |a_2|, \cdots, |a_N|, |a_N| + \epsilon\}\). Then \(\forall n\in\N[|a_n| \le M]\). □

Proof Let \(A_n = \{a_m| m\ge n\}\). Proof 1. Since \(a_n \in A_n\), by definition, \(\inf A_n \le a_n \le \sup A_n\). Then \(l_n \le a_n \le u_n\) for all \(n\in\N\).

Proof 2. \(A_{n+1} = \{a_m| m\ge n+1\}\), then \(A_n = A_{n+1}\cap \{a_n\}\). If \(a_n < \inf A_{n+1}\), then \(\inf A_n \le a_n \le \inf A_{n+1}\). Else if \(a_n > \inf A_{n+1}\), then \(\inf A_n = \inf A_{n+1}\). Therefore, \(l_n\) is nondecreasing.

The proof of \(u_n\) is similar. □

Proof Let \(A_n = \{a_m|m\ge n\}, B_n = \{b_m|m\ge n\}, C_n = \{a_m + b_m|m\ge n\}\).

\(\forall n,m\in\N, m\ge n[a_m \le\sup A_n \land b_m\le\sup B_n \Rarr a_m + b_m \le\sup A_n + \sup B_n]\), i.e., \(\sup A_n + \sup B_n\) is an upper bound of \(C_n\). Thus \(\sup C_n \le \sup A_n + \sup B_n\).

Therefore, \(\varlimsup_{n\to\infty}(a_n + b_n) = \lim_{n\to\infty} C_n \le \lim_{n\to\infty} A_n + \lim_{n\to\infty} B_n = \varlimsup_{n\to\infty} a_n + \varlimsup_{n\to\infty} b_n\). □

Proof Let \(A_n = \{a_m| m\ge n\}\).

(\(\Rarr\)).

\(a_n\) converges, then \(\forall \epsilon > 0 \exists n \in\N [n\ge N \Rarr |a_n - L| < \epsilon \Lrarr L - \epsilon < a_n < L + \epsilon]\).

\(a_n \in A_n \Rarr L - \epsilon < a_n \le \sup A_n = u_n\).

And \(L + \epsilon\) is an upper bound of \(A_n\), then \(u_n = \sup A_n \le L + \epsilon\).

Therefore, \(|u_n - L| < \epsilon\), i.e., \(u_n\) converges to \(L\) as \(a_n\) does.

The proof of \(l_n\) is similar.

(\(\Larr\)).

It is obviously true by the squeeze test. □

Proof (\(\Rarr\)). Already proved.

(\(\Larr\)).

Assume that \(a_n\) is Cauchy. We claim that \(\lim_{n\to\infty}(u_n - l_n) = 0\).

For \(\epsilon > 0\), there is some \(N\in\N\) such that \(m,n \ge N \Rarr |a_n - a_m| < \epsilon\).

In particular,

\begin{align*} n \ge N & \Rarr a_N - \epsilon < a_n < a_N +\epsilon\\ &\Rarr a_N - \epsilon \le l_N \le u_N \le a_N + \epsilon \\ &\Rarr 0 \le u_n - l_n \le u_N - l_N \le 2\epsilon \end{ali□

Proof Proof 1. \(\forall s, s_1\in S[|s - s_1| \le \epsilon \Rarr s_1 - \epsilon \le s \le s_1 + \epsilon]\).

Thus, \(s_1 - \epsilon, s_1 +\epsilon\) is the lower bound and upper bound of \(S\) respectively.

Let \(M = \max\{|s_1 - \epsilon|, |s_1 +\epsilon|\}\). Then \(\forall s\in S[|s|\le M]\).

Proof 2. Assume that \(\sup S - \inf S > \epsilon\), i.e., \(\sup S > \inf S + \epsilon\), \(\inf S + \epsilon\) is not an upper bound of \(S\), and hence there must be a \(s_2 \in S\), such that \(\inf S + \epsilon \le s_2\), then \(\inf S \le s_2 - \epsilon\).

Since \(|s - s_2| \le \epsilon\), like Proof 1, \(s_2 - \epsilon\) is a lower bound of \(S\), i.e., \(s_2 - \epsilon \le \inf S\), a contradiction.

Thus \(\sup S - \inf S \le \epsilon\). □