Proof. Let \(\varepsilon > 0\) be arbitrary. Since \({(b_n)}_{n \in {\mathbb{N}}}\) is bounded, there exists a real number \(M > 0\) such that \(|b_n| \le M\) for all \(n \in {\mathbb{N}}.\) Because of the convergence \({(a_n)}_{n \in {\mathbb{N}}},\) for \(\widetilde{\varepsilon} := \frac{\varepsilon}{M}\) there exists an \(N \in {\mathbb{N}}\) with \(|a_n| \le \frac{\varepsilon}{M}\) for all \(n \ge N.\) (For simplicity, from now on we will drop the dependency of \(N\) on \(\varepsilon\) in the notation and simply write \(N\) instead of \(N_{\widetilde{\varepsilon}}.\)) Then we obtain \[|a_nb_n - 0| = |a_n| \cdot |b_n| \le \frac{\varepsilon}{M} \cdot M = \varepsilon\] for all \(n \ge N.\) Since \(\varepsilon > 0\) is arbitrary, it follows that \(\lim_{n \to \infty} a_nb_n = 0.\)

Proof.

1. Let \(\varepsilon > 0\) be arbitrary. Then for \(\widetilde{\varepsilon} = \frac{\varepsilon}{2}\) there exist \(\widehat{N},\,\widetilde{N} \in {\mathbb{N}}\) such that \[\begin{aligned} |a_n - a| &< \frac{\varepsilon}{2} \quad \text{for all } n \ge \widehat{N}, \\ |b_n - b| &< \frac{\varepsilon}{2} \quad \text{for all } n \ge \widetilde{N}. \end{aligned}\] Now set \(N := \max\big\{ \widehat{N},\,\widetilde{N} \big\}.\) Then by the triangle inequality, for all \(n \ge N\) we have \[|(a_n+b_n) - (a+b)| \le |a_n - a| + |b_n - b| \le \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.\] Since \(\varepsilon > 0\) is arbitrary, it follows that the limit of \({(a_n + b_n)}_{n \in {\mathbb{N}}}\) is \(a+b.\)

2. Exercise!

3. Since the sequence \({(a_n-a)}_{n \in {\mathbb{N}}}\) is a null sequence and since \({(b_n)}_{n \in {\mathbb{N}}}\) is bounded by Theorem 2.2, the sequence \({((a_n-a)b_n)}_{n \in {\mathbb{N}}} = {(a_nb_n-ab_n)}_{n \in {\mathbb{N}}}\) is also a null sequence by Theorem 2.3. Together with i and ii we obtain that the sequence \({(a_nb_n)}_{n \in {\mathbb{N}}} = {((a_nb_n - ab_n) + ab_n)}_{n \in {\mathbb{N}}}\) converges. Moreover, we have \[\lim_{n \to \infty} a_n b_n = \lim_{n \to \infty}(a_nb_n - ab_n) + \lim_{n \to \infty} ab_n = 0 + ab = ab.\]

4. For the proof of this claim, it is sufficient to show that \(\big(\frac{1}{b_n} \big)_{n \in {\mathbb{N}}}\) converges towards \(\frac{1}{b},\) since then we can apply iii to the sequence \(\big(a_n \cdot \frac{1}{b_n} \big)_{n \in {\mathbb{N}}}.\) We have \[\frac{1}{b_n} - \frac{1}{b} = \frac{b-b_n}{b_nb} = (b-b_n) \frac{1}{b_nb}.\] Since \({(b-b_n)}_{n \in {\mathbb{N}}}\) is a null sequence, it remains to show that \(\big(\frac{1}{b_nb}\big)_{n \in {\mathbb{N}}}\) is bounded, because then also \(\big(\frac{1}{b_n} - \frac{1}{b}\big)_{n \in {\mathbb{N}}}\) is a null sequence by Theorem 2.3. Since \({(b_n)}_{n \in {\mathbb{N}}}\) converges towards \(b,\) for \(\varepsilon := \frac{|b|}{2}\) there exists an \(N \in {\mathbb{N}}\) such that \(|b_n-b| < \frac{|b|}{2}\) for all \(n \ge N.\) With this we obtain \[\begin{aligned} |b_n| = |b_n| + \frac{|b|}{2} - \frac{|b|}{2} &> |b_n| + |b-b_n| - \frac{|b|}{2} \\ &\ge |b_n + b - b_n| - \frac{|b|}{2} = \frac{|b|}{2}. \end{aligned}\] From that we obtain \[\frac{1}{|b_n|} < \frac{2}{|b|}, \quad \text{resp.} \quad \frac{1}{|b_nb|} < \frac{2}{|b|^2}\] for all \(n \ge N\) and therefore, \(\big(\frac{1}{b_nb}\big)_{n \in {\mathbb{N}}}\) is bounded. Hence, as desired, \(\big(\frac{1}{b_n} - \frac{1}{b}\big)_{n \in {\mathbb{N}}}\) is a null sequence and \(\big(\frac{1}{b_n} \big)_{n \in {\mathbb{N}}}\) converges towards \(\frac{1}{b}.\)

Proof. Let \({(a_n)}_{n \in {\mathbb{N}}}\) be monotonic and bounded. Without loss of generality, let \({(a_n)}_{n \in {\mathbb{N}}}\) be monotonically increasing, the other case is completely analogous. Then, the set \[M: = \{a_n \; | \; n \in {\mathbb{N}}\}\] is bounded from above. By the completeness axiom (Axiom 1.3), there exists \(a:= \sup M.\) We show now that \({(a_n)}_{n \in {\mathbb{N}}}\) converges towards \(a\): Therefore, let \(\varepsilon > 0\) be arbitrary. By Theorem 1.2 there exists an \(x \in M\) (or an \(N \in {\mathbb{N}}\) with \(x = a_N\)) and \(a-\varepsilon < a_N \le a.\) Because of the monotonicity of the sequence we obtain \[a-\varepsilon < a_N \le a_n \le a < a+\varepsilon \quad \text{for all } n \ge N.\] But this means that \(|a_n - a| < \varepsilon\) for all \(n \ge N.\) Hence, \(\lim_{n \to \infty} a_n = a,\) because \(\varepsilon > 0\) is arbitrary.

Proof. If \({(a_n)}_{n \in {\mathbb{N}}}\) converges and the limit \(a\) is positive, then with Theorem 2.4 we obtain \[a = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \frac{1}{2} \left(a_n+ \frac{c}{a_n} \right) = \frac{1}{2}a + \frac{1}{2} \frac{c}{a} \quad \Longrightarrow \quad \frac{1}{2}a = \frac{1}{2} \frac{c}{a}.\] This finally gives \(a^2 = c.\) Let us show that \(a\) is the unique positive solution of the equation \(x^2 = c\): Assume that \(b > 0\) also satisfies \(b^2 = c.\) Then \[0 = a^2 - b^2 = (a+b)(a-b) \quad \Longrightarrow \quad a+b = 0 \text{ or } a-b = 0.\] Since \(a,\,b > 0,\) we must have \(a-b = 0,\) hence \(a=b\) and the square root is unique.

We still have to prove that \({(a_n)}_{n \in {\mathbb{N}}}\) indeed converges to a positive limit. By induction one can show that \(a_0 > 0\) implies that \(a_n > 0\) for all \(n \in {\mathbb{N}}\) (exercise!). At this moment this only implies that the sequence \({(a_n)}_{n \in {\mathbb{N}}}\) is well-defined. (Note that there would be a problem if one of the elements \(a_n\) is zero.) Further we have

1. The sequence \({(a_n)}_{n \ge 1}\) is monotonically decreasing, since for all \(n \ge 1\) we have \[a_n - a_{n+1} = a_n - \frac{1}{2} \left( a_n + \frac{c}{a_n} \right) = \frac{1}{2}a_n - \frac{1}{2a_n}c = \frac{1}{2a_n} \big(a_n^2 - c\big) \ge 0.\] The latter inequality follows from \[\begin{aligned} a_n^2 - c &= \frac{1}{4} \left( a_{n-1} + \frac{c}{a_{n-1}} \right)^2 - c \\ &= \frac{1}{4} \left( a_{n-1}^2 + 2c+ \frac{c^2}{a_{n-1}^2} - 4c \right) \\ &= \frac{1}{4} \left( a_{n-1} - \frac{c}{a_{n-1}} \right)^2 \ge 0. \end{aligned}\]

2. The sequence \({(a_n)}_{n \in {\mathbb{N}}}\) is bounded, because by the monotonicity we have \[0 < a_n \le a_1 \quad \text{for all } n \ge 1.\]

Together with the monotone convergence theorem Theorem 2.8, the sequence \({(a_n)}_{n \ge 1}\) converges, hence also \({(a_n)}_{n \in {\mathbb{N}}}\) converges and the limit \(a\) exists.

It still remains to show that \(a > 0.\) (This does not follow from \(a_n > 0\) for all \(n \in {\mathbb{N}}\)!). First, by Corollary 2.6 we get \(a \ge 0.\) On the other hand, from i we obtain \[a^2 = \lim_{n \to \infty} a_n^2 \ge c > 0,\] hence also \(a > 0.\)

Proof. Let \({(a_n)}_{n \in {\mathbb{N}}}\) be a convergent sequence of real numbers with limit \(a \in {\mathbb{R}}\) and let \(\varepsilon > 0\) be arbitrary. Then for \(\widetilde{\varepsilon} := \frac{\varepsilon}{2},\) there exists an \(N \in {\mathbb{N}}\) such that \(|a_n - a| < \frac{\varepsilon}{2}.\) By applying the triangle inequality we obtain for \(n,\,m \ge N\) that \[|a_n - a_m| = |a_n - a + a - a_m| \le |a_n-a| + |a - a_m| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.\] Since \(\varepsilon > 0\) is arbitrary, this shows that \({(a_n)}_{n \in {\mathbb{N}}}\) is a Cauchy sequence.

Proof. Consider the set \[M:=\left\{ n \in {\mathbb{N}}\; | \; a_n = \max\{ a_m \; | \; m \ge n \}\right\}=\left\{ n \in {\mathbb{N}}\; | \; a_n \ge a_m \text{ for all } m\ge n\right\},\] i. e., the set of indices of those sequence elements which do not have a strictly larger successor. We distinguish two cases:
Case 1: \(M\) contains infinitely elements. Then those \(n_k \in {\mathbb{N}}\) with \(n_k \in M\) form a strictly monotonically increasing sequence \({(n_k)}_{k \in {\mathbb{N}}}.\) Thus, \({(a_{n_k})}_{k \in {\mathbb{N}}}\) is a subsequence of \({(a_n)}_{n \in {\mathbb{N}}}\) and because of \(a_{n_k} \ge a_{n_{k+1}}\) for all \(k \in {\mathbb{N}}\) (since \(n_k \in M\)), the subsequence \({(a_{n_k})}_{k \in {\mathbb{N}}}\) is monotonically decreasing.
Case 2: \(M\) contains finitely many elements. Define \[n_0 := \begin{cases} \max M + 1, & \text{if } M \neq \emptyset, \\ 0, & \text{if } M = \emptyset. \end{cases}\] Since \(n_0 \notin M,\) there exists an \(n_1 > n_0\) such that \(a_{n_0} < a_{n_1}.\) But since also \(n_1 \notin M,\) there exists an \(n_2 > n_1\) with \(a_{n_1} < a_{n_2}.\) Inductively, we can construct a subsequence \({(a_{n_k})}_{k \in {\mathbb{N}}}\) such that \[a_{n_0} < a_{n_1} < a_{n_2} < \ldots,\] i. e., a strictly monotonically increasing subsequence.

Proof. By Lemma 2.11, \({(a_n)}_{n \in {\mathbb{N}}}\) has a monotonic subsequence \({(a_{n_k})}_{k \in {\mathbb{N}}}.\) If \({(a_n)}_{n \in {\mathbb{N}}}\) is bounded then also \({(a_{n_k})}_{k \in {\mathbb{N}}}\) is clearly bounded. Hence, \({(a_{n_k})}_{k \in {\mathbb{N}}}\) is monotonic and bounded, hence it is convergent by Theorem 2.8.

Proof. Let \({(a_n)}_{n \in {\mathbb{N}}}\) be a Cauchy sequence and let \(a \in {\mathbb{R}}\) be an accumulation point of \({(a_n)}_{n \in {\mathbb{N}}}.\) Then by definition there exists a subsequence \({(a_{n_k})}_{k \in {\mathbb{N}}}\) of \({(a_n)}_{n \in {\mathbb{N}}}\) with \(\lim_{k \to \infty} a_{n_k} = a.\) Let \(\varepsilon > 0\) be arbitrary. Then there exists an \(\widetilde{N} \in {\mathbb{N}}\) such that \[\big|a_{n_k} - a\big| < \frac{\varepsilon}{2} \quad \text{for all } k\ge \widetilde{N}.\] Since \({(a_n)}_{n \in {\mathbb{N}}}\) is a Cauchy sequence, there exists an \(\widehat{N} \in {\mathbb{N}}\) such that \[|a_n - a_m| < \frac{\varepsilon}{2} \quad \text{for all } n,\,m \ge \widehat{N}.\] Set \(N:= \max\big\{ \widetilde{N},\,\widehat{N} \big\}.\) Then for all \(n,\,k \ge N\) we get \[|a_n - a| \le \big|a_n - a_{n_k}\big| + \big|a_{n_k} - a\big| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.\] Since \(\varepsilon > 0\) is arbitrary, \({(a_n)}_{n \in {\mathbb{N}}}\) converges towards \(a.\)

Proof. Let \({(a_n)}_{n \in {\mathbb{N}}}\) be a Cauchy sequence. Analogously to the proof of Theorem 2.2, one can show that \({(a_n)}_{n \in {\mathbb{N}}}\) is bounded (exercise!). Then by Theorem 2.12, \({(a_n)}_{n \in {\mathbb{N}}}\) has a convergent subsequence, i. e., it has an accumulation point. Now by Lemma 2.13, the sequence \({(a_n)}_{n \in {\mathbb{N}}}\) is convergent.

Proof. We prove this results using Cauchy’s convergence criterion (Theorem 2.14). First we show that \({(a_n)}_{n \in {\mathbb{N}}}\) is a Cauchy sequence. Let \(\varepsilon > 0\) be arbitrary. Since \({(b_n - a_n)}_{n \in {\mathbb{N}}}\) is a null sequence, there exists an \(N \in {\mathbb{N}}\) such that \[b_n - a_n = |b_n - a_n| < \varepsilon \quad \text{for all } n \ge N.\] Since \(I_m \subseteq I_n\) for all \(m \ge n,\) we obtain \(a_n \le a_m \le b_n.\) Therefore, for all \(m \ge n \ge N\) we obtain \[|a_m - a_n| = a_m - a_n \le b_n - a_n < \varepsilon.\] This shows that \({(a_n)}_{n \in {\mathbb{N}}}\) is a Cauchy sequence. Similarly, one can show that also \({(b_n)}_{n \in {\mathbb{N}}}\) is a Cauchy sequence. Thus there exist \(a,\,b \in {\mathbb{R}}\) such that \[a:= \lim_{n \to \infty} a_n,\quad b:=\lim_{n \to \infty} b_n.\] Both limits are identical because \[0 = \lim_{n \to \infty} (b_n -a_n) = \lim_{n \to \infty} b_n - \lim_{n \to \infty} a_n = b-a,\] hence \(a=b.\) Since \(a_n < b_n\) for all \(n \in {\mathbb{N}},\) an application of Corollary 2.6 yields \(a \le b_n\) but also \(a_n \le b\) for all \(n \in {\mathbb{N}}.\) Hence we obtain \(a_n \le b = a \le b_n\) for all \(n \in {\mathbb{N}}.\) Hence \(a \in I_n\) for all \(n \in {\mathbb{N}}\) and so \(\bigcap_{n \in {\mathbb{N}}} I_n \supseteq \{a\}.\) The other inclusion \(\bigcap_{n \in {\mathbb{N}}} I_n \subseteq \{a\}\) is then also clear.

Proof. Let \(g_k = \sup_{n \ge k} a_n\) for \(k \in {\mathbb{N}}.\) Next we show the equivalence in the theorem.
“\(\Longrightarrow\)”: Let \(\varepsilon > 0\) be arbitrary. Since \(a = \limsup_{n \to \infty} a_n = \lim_{k \to \infty}g_k\) there exists an \(N \in {\mathbb{N}}\) such that \[a-\varepsilon < g_k < a+\varepsilon \quad \text{for all } k \ge N.\] Since \(a_n \le g_N\) for all \(n \ge N,\) the condition in i follows. Assume ii would not be true. Then \(a_n > a - \varepsilon\) would only hold for finitely many \(n \in {\mathbb{N}}.\) Hence, there exists an \(\widetilde{N} \in {\mathbb{N}}\) such that \(a_n \le a - \varepsilon\) for all \(n \ge \widetilde{N}.\) Thus, also \(g_k \le a-\varepsilon\) for all \(k \ge \widetilde{N}.\) With Corollary 2.6 we obtain the contradiction \(a = \lim_{k \to \infty} g_k \le a - \varepsilon.\) Hence, also ii is satisfied.
“\(\Longleftarrow\)”: Let \(\varepsilon > 0.\) With ii it holds that \(g_k > a - \varepsilon\) for all \(k \in {\mathbb{N}}\) and hence, \(\lim_{k \to \infty} g_k \ge a - \varepsilon.\) Then with Remark 2.6 we get \[\lim_{k \to \infty} g_k \ge a.\] On the other hand, by i, for every \(\varepsilon > 0\) there exists an \(N \in {\mathbb{N}}\) with \(a_n < a+ \varepsilon\) for all \(n \ge N.\) Hence we have \(g_k \le a + \varepsilon\) for all \(k \ge N.\) Then we also get \(\lim_{k \to \infty} g_k \le a + \varepsilon\) and with Remark 2.6 we obtain \[\lim_{k \to \infty} g_k \le a.\] Summarizing we see that \[\limsup_{n \to \infty} a_n = \lim_{k \to \infty} g_k = a.\]