3. Sequences

Let \(\varepsilon \gt 0\) be arbitrary but fixed. By the Archimedean property of \(\real\), there exists \(K\in\N\) such that \(\frac{1}{K} \lt \varepsilon\). Then, if \(n\geq K\) then \(\frac{1}{n}\leq \frac{1}{K} \lt \eps\). Therefore, if \(n\geq K\) then \begin{align*} \left|x_n - 0\right| &= \left|\tfrac{1}{n}- 0\right|\\ &= \frac{1}{n}\\ & \leq \frac{1}{K}\\ & \lt \eps. \end{align*} This proves, by definition, that \(\limi \frac{1}{n} = 0\).

Given an arbitrary \(\varepsilon \gt 0\), we want to prove that there exists \(K\in\N\) such that \[ \left|\frac{3n+2}{n+1} - 3\right| \lt \varepsilon,   \forall\; n\geq K. \] Start by analyzing \(|x_n-L|\): \begin{align*} |x_n-L| &=\left|\frac{3n+2}{n+1} - 3\right|\\ & = \left|\frac{-1}{n+1}\right|\\ & = \frac{1}{n+1}. \end{align*} Now, the condition that \[ \left|\frac{3n+2}{n+1} - 3\right| = \frac{1}{n+1} \lt \varepsilon \] it is equivalent to \[ \frac{1}{\varepsilon} - 1 \lt n. \] Now let's write the formal proof. Let \(\varepsilon \gt 0\) be arbitrary and let \(K\in\N\) be such that \(\frac{1}{\varepsilon} - 1 \lt K\). Then, \(\frac{1}{K+1} \lt \varepsilon\). Now if \(n\geq K\) then \(\frac{1}{n+1} \leq \frac{1}{K+1}\) and thus if \(n\geq K\) then \begin{align*} \left|\frac{3n+2}{n+1} - 3\right| &= \left|\frac{-1}{n+1}\right|\\ &= \frac{1}{n+1}\\ & \leq \frac{1}{K+1}\\ & \lt \varepsilon. \end{align*} By definition, this proves that \((x_n) \rightarrow 3\).

Let \(x_n = \frac{4n^3+3n}{n^3+6} \). We want to show that for any given \(\varepsilon \gt 0\), there exists \(K\in\N\) such that if \(n\geq K\) then \[ |x_n-4|=\left|\frac{4n^3+3n}{n^3+6} - 4\right| \lt \varepsilon. \] Start by analyzing \(|x_n-4|\): \begin{align*} |x_n-4|&=\left|\frac{4n^3+3n}{n^3+6} - 4\right| \\ &= \left|\frac{3n-24}{n^3+6}\right| \end{align*} In this case, it is difficult to explicitly isolate for \(n\) in terms of \(\varepsilon\). Instead we take a different approach; we find an upper bound for \(|x_n-4|=\left|\frac{3n-24}{n^3+6}\right|\): \begin{align*} \left|\frac{3n-24}{n^3+6}\right| &\leq \frac{3n+24}{n^3+6}\\[2ex] & \leq \frac{27n}{n^3 + 6}\\[2ex] & \lt \frac{27n}{n^3} \\ &= \frac{27}{n^2}. \end{align*} Hence, if \(\frac{27}{n^2} \lt \varepsilon\) then also \(|x_n-4| \lt \varepsilon\) by the transitivity property of inequalities. The inequality \(\frac{27}{n^2} \lt \varepsilon\) holds true if and only if \(\sqrt{\frac{27}{\varepsilon}} \lt n\). Now that we have done a detailed preliminary analysis, we can proceed with the proof. Suppose that \(\varepsilon \gt 0\) is given and let \(K\in\N\) be such that \(\sqrt{\frac{27}{\varepsilon}} \lt K \). Then \(\frac{27}{\varepsilon} \lt K^2\), and thus \(\frac{27}{K^2} \lt \varepsilon\). Then, if \(n\geq K\) then \(\frac{27}{n^2}\leq \frac{27}{K^2}\) and therefore \begin{align*} |x_n -4| &= \left|\frac{3n-24}{n^3+6}\right|\\[2ex] & \leq \frac{3n+24}{n^3+6}\\ & \leq \frac{27n}{n^3 + 6}\\ & \lt \frac{27n}{n^3}\\ & = \frac{27}{n^2}\\ &\leq \frac{27}{K^2}\\ & \lt \varepsilon. \end{align*} This proves that \(\limi x_n = 4\).

Let \((\delta_n)\) be any sequence of positive numbers converging to zero, for example, \(\delta_n = \frac{1}{n}\). Now since \(\zeta-\delta_n \lt \zeta+\delta_n\) for each \(n\in\N\), then by the Density theorem there exists a rational number \(x_n\) such that \(\zeta-\delta_n \lt x_n \lt \zeta+\delta_n\). In other words, \(|x_n-\zeta| \lt \delta_n\). Now let \(\eps \gt 0\) be arbitrary. Since \((\delta_n)\) converges to zero, there exists \(K\in\N\) such that \(|\delta_n - 0| \lt \eps\) for all \(n\geq K\), or since \(\delta_n \gt 0\), then \(\delta_n \lt \eps\) for all \(n\geq K\). Therefore, if \(n\geq K\) then \(|x_n-\zeta| \lt \delta_n \lt \eps\). Thus, for arbitrary \(\eps \gt 0\) there exists \(K\in\N\) such that if \(n\geq K\) then \(|x_n-\zeta| \lt \eps\). This proves that \((x_n)\) converges to \(\zeta\).

We want to prove given any \(\varepsilon \gt 0\) there exists \(K\in\N\) such that \[ \left|\frac{\cos(n)}{n^2-1}\right| \lt \varepsilon,  n\geq K. \] Now, since \(|\cos(x)|\leq 1\) for all \(x\in\real\) we have that \begin{align*} \left|\frac{\cos(n)}{n^2-1}\right| &= \frac{|\cos(n)|}{n^2-1}\\ & \leq \frac{1}{n^2-1}\\ & \leq \frac{1}{n^2-\frac{1}{2}n^2}\\ & = \frac{2}{n^2}. \end{align*} Thus, if \(\frac{2}{n^2} \lt \varepsilon\) then \(\left|\frac{\cos(n)}{n^2-1}\right|\). Let \(\varepsilon \gt 0\) be arbitrary. Let \(K\in\N\) be such that \(\sqrt{\frac{2}{\varepsilon}} \lt K\). Then \(\frac{2}{K^2} \lt \varepsilon\). Therefore, if \(n\geq K\) then \(\frac{2}{n^2}\leq \frac{2}{K^2}\) and therefore \begin{align*} \left|\frac{\cos(n)}{n^2-1}\right| &= \frac{|\cos(n)|}{n^2-1}\\ & \leq \frac{1}{n^2-1}\\ &\leq \frac{1}{n^2-\frac{1}{2}n^2}\\ & = \frac{2}{n^2} \leq \frac{2}{K^2}\\ & \lt \varepsilon \end{align*} This proves that \(\limi x_n = 0\).

Let \(\varepsilon \gt 0\) be arbitrary. Since \(a_n\rightarrow 0\), there exists \(K_1\in\N\) such that \(a_n \lt \varepsilon\) for all \(n\geq K_1\). Let \(K=\max\{M, K_1\}\). Then, if \(n\geq K\) then \(a_n \lt \varepsilon\) and \(|x_n-L|\leq a_n\). Thus, if \(n\geq K\) then \[ |x_n - L|\leq a_n \lt \varepsilon. \]

We first note that \[ r = \frac{1}{\tfrac{1}{r}} = \frac{1}{1 + x} \] where \(x=\frac{1}{r}-1\) and since \(r \lt 1\) then \(x \gt 0\). Now, by Bernoulli's inequality (Example 1.2.5) it holds that \((1+x)^n \geq 1 + x n \) for all \(n\in \N\) and therefore \begin{align*} r^n &= \frac{1}{\left(1 + x\right)^n}\\ & \leq \frac{1}{1+nx}\\ & \lt \frac{1}{n x}. \end{align*} Now since \(\lim_{n\rightarrow\infty} \frac{1}{n x} = 0\) then it follows by Theorem 3.1.9 that \[ \limi r^n = 0. \]

We have that \begin{align*} \left| \frac{n^2-1}{2n^2+3}-\frac{1}{2} \right| &= \frac{5}{2}\frac{1}{(2n^2+3)}\\ & \lt \frac{5/2}{2n^2}. \end{align*} Using the definition of the limit of a sequence, one can show that \(\limi \frac{5}{4n^2} = 0\) and therefore \(\limi x_n = \frac{1}{2}\).

Suppose that \((x_n)\rightarrow L_1\) and that \((x_n)\rightarrow L_2\). Let \(\varepsilon \gt 0\) be arbitrary. Then there exists \(K_1\) such that \(|x_n-L_1| \lt \varepsilon/2\) for all \(n\geq K_1\) and there exists \(K_2\) such that \(|x_n-L_2| \lt \varepsilon/2\) for all \(n\geq K_2\). Let \(K=\max\{K_1,K_2\}\). Then for \(n\geq K\) it holds that \(|x_n-L_1| \lt \eps/2\) and also \(|x_n-L_2| \lt \eps/2\) and therefore \begin{align*} |L_1-L_2| &= |L_1-x_n+x_n-L_2|\\ & \lt |x_n-L_1| + |x_n-L_2|\\ & \lt \varepsilon/2+\varepsilon/2\\ & = \varepsilon. \end{align*} Hence, \(|L_1-L_2| \lt \varepsilon\) for all \(\varepsilon \gt 0\), and therefore by Theoreom 2.2.7 we conclude that \(|L_1-L_2|=0\), that is, \(L_1-L_2=0\).

Suppose that \((x_n)\) converges to \(L\). Then there exists \(K\in\N\) such that \(|x_n-L| \lt 1\) for all \(n\geq K\). Let \[ R = 1 + \max\{|x_1-L|,|x_2-L|,\ldots,|x_{K-1}-L|\}, \] and we note that \(R\geq 1\). Then for all \(n\geq 1\) it holds that \(|x_n - L| \leq R\). Indeed, if \(n\geq K\) then \(|x_n-L| \lt 1\leq R\) and if \(1\leq n\leq K-1\) then \[ |x_n - L|\leq \max\{|x_1-L|, \ldots,|x_{K-1}-L|\} \leq R. \] Thus, for all \(n\geq 1\) it holds that \(L-R \leq x_n \leq R+L\) and this proves that \((x_n)\) is bounded.

Follows by the inequality \(||x_n|-|L||\leq |x_n-L|\) (see Corollary 2.3.6). Indeed, for any given \(\eps \gt 0\) there exists \(K\in\N\) such that \(|x_n-L| \lt \eps\) for all \(n\geq K\) and therefore \(||x_n|-|L||\leq |x_n - L| \lt \eps\) for all \(n\geq K\).

(i) By the triangle inequality \begin{align*} |x_n+y_n-(L+M)| &= |x_n-L + y_n-M|\\ & \lt |x_n-L| + |y_n-M|. \end{align*} Let \(\varepsilon \gt 0\). There exists \(K_1\) such that \(|x_n-L| \lt \varepsilon/2\) for \(n\geq K_1\) and there exists \(K_2\) such that \(|y_n-M| \lt \varepsilon/2\) for \(n\geq K_2\). Let \(K=\max\{K_1,K_2\}\). Then for \(n\geq K\) \begin{align*} |x_n+y_n-(L+M)| & \leq |x_n-L| + |y_n-M| \\ & \lt \varepsilon/2 + \varepsilon/2\\ &= \varepsilon. \end{align*} The proof for \((x_n-y_n)\rightarrow L-M\) is similar.
(ii) We have that \begin{align*} |x_ny_n-LM| &= |x_ny_n - y_nL + y_nL - LM|\\ &\leq |x_ny_n-y_nL| + |y_nL-LM|\\ &= |y_n||x_n-L| + |L||y_n-M|. \end{align*} Now, \((y_n)\) is bounded because it is convergent, and therefore \(|y_n|\leq R\) for all \(n\in\N\) for some \(R \gt 0\). By convergence of \((y_n)\) and \((x_n)\), there exists \(K\in\N\) such that \(|x_n-L| \lt \frac{\varepsilon}{2R}\) and \(|y_n-M| \lt \frac{\varepsilon}{2(|L|+1)}\) for all \(n\geq K\). Therefore, if \(n\geq K\) then \begin{align*} |x_ny_n-LM| & \lt |y_n||x_n-L| + |L||y_n-M|\\[2ex] & \lt R |x_n-L| + (|L|+1)|y_n-M|\\[2ex] & \lt R\frac{\varepsilon}{2R} + (|L|+1)\frac{\varepsilon}{2(|L|+1)}\\[2ex] &= \varepsilon. \end{align*} (iii) It is enough to prove that \(\left(\frac{1}{y_n}\right)\rightarrow\frac{1}{M}\) and then use (ii). Now, since \(M\neq 0\) and \(y_n\neq 0\) then \(|y_n|\) is bounded below by some positive number, say \(R \gt 0\). Indeed, \((|y_n|)\rightarrow |M|\) and \(|y_n| \gt 0\). Thus, \(\frac{1}{|y_n|} \lt \frac{1}{R}\) for all \(n\in\N\). Now, \begin{align*} \left|\frac{1}{y_n}-\frac{1}{M}\right| &= \frac{1}{|y_n||M|}|y_n-M|\\[2ex] & \lt \frac{1}{R|M|} |y_n-M|. \end{align*} For \(\varepsilon \gt 0\), there exists \(K\in\N\) such that \(|y_n-M| \lt R|M|\varepsilon\) for all \(n\geq K\). Therefore, for \(n\geq K\) we have that \begin{align*} \left|\frac{1}{y_n}-\frac{1}{M}\right| & \lt \frac{1}{R|M|} |y_n-M|\\[2ex] & \lt \frac{1}{R|M|} R|M|\varepsilon\\[2ex] &= \varepsilon. \end{align*}

We prove the contrapositive, that is, we prove that if \(L \lt 0\) then there exists \(K\in\N\) such that \(x_K \lt 0\). Suppose then that \(L \lt 0\). Let \(\varepsilon \gt 0\) be such that \(L+\varepsilon \lt 0\). Since \((x_n)\rightarrow L\), there exists \(K\in\N\) such that \(x_K \lt L+\varepsilon\), and thus by transitivity we have \(x_K \lt 0\).

Suppose for now that \(M=1\), that is, \(x_n\leq y_n\) for all \(n\in\N\). Consider the sequence \(z_n = y_n-x_n\). Then \(z_n\geq 0\) for all \(n\in \N\) and \((z_n)\) is convergent since it is the difference of convergent sequences. By Theorem 3.2.7, we conclude that \(\limi z_n \geq 0\). But \begin{align*} \limi z_n &= \limi(y_n-x_n)\\ & = \limi y_n -\limi x_n \end{align*} and therefore \(\limi y_n - \limi x_n \geq 0\), which is the same as \(\limi y_n \geq \limi x_n\). If \(M \gt 1\), then we can apply the theorem to the \(M\)-tail of the sequences of \((x_n)\) and \((y_n)\) and the result follows.

We have that \(0\leq x_n -a \leq b-a\). The sequence \(y_n = b-a\) is constant and converges to \(b-a\). The sequence \(z_n = x_n - a\) converges to \(L-a\). Therefore, by the previous theorem, \(0\leq L-a\leq b-a\), or \(a\leq L\leq b\).

Let \(\varepsilon \gt 0\) be arbitrary. There exists \(K_1\in\N\) such that \(L-\varepsilon \lt y_n \lt L+\varepsilon\) for all \(n\geq K_1\) and there exists \(K_2\in\N\) such that \(L-\varepsilon \lt z_n \lt L+\varepsilon\) for all \(n\geq K_2\). Let \(K=\max\{K_1,K_2\}\). Then if \(n\geq K\) then \[ L-\varepsilon \lt y_n \leq x_n \leq z_n \lt L+\varepsilon. \] Therefore, for \(n\geq K\) we have that \[ L-\varepsilon \lt x_n \lt L+\varepsilon \] and thus \(\limi x_n = L\).

Let \(r\in\real\) be such that \(L \lt r \lt 1\) and set \(\varepsilon = r-L\). There exists \(K\in\N\) such that \[ \frac{x_{n+1}}{x_n} \lt L+\varepsilon = r \] for all \(n\geq K\). Therefore, for all \(n\geq K\) we have that \[ 0 \lt x_{n+1} \lt r x_n . \] Thus, \(x_{K+1} \lt r x_K\), and therefore \(x_{K+2} \lt r x_{K+1} \lt r^2 x_K\), and inductively for \(m\geq 1\) it holds that \[ x_{K+m} \lt r^m x_K. \] Hence, the tail of the sequence \((x_n)\) given by \((y_m) = (x_{K+1}, x_{K+2}, \ldots,)\) satisfies \[ 0 \lt y_m \lt r^m x_K. \] Since \(0 \lt r \lt 1\) it follows that \(\lim_{m\rightarrow\infty} r^m = 0\) and therefore \(\lim_{m\rightarrow\infty}y_m = 0\) by the Squeeze theorem. This implies that \((x_n)\) converges to \(0\) also.

Suppose that \((x_n)\) is bounded above and increasing. Let \(u=\sup\{x_n\;|\;n\in\N\}\) and let \(\varepsilon \gt 0\) be arbitrary. Then by the properties of the supremum, there exists \(x_K\) such that \(u-\varepsilon \lt x_K\leq u\). Since \((x_n)\) is increasing, and \(u\) is an upper bound for the range of the sequence, it follows that \(x_K\leq x_n\leq u\) for all \(n\geq K\). Therefore, \(u-\varepsilon \lt x_n\leq u\) for all \(n\geq K\). Clearly, this implies that \(u-\varepsilon \lt x_n \lt u+\varepsilon\) for all \(n\geq K\). Since \(\varepsilon \gt 0\) was arbitrary, this proves that \((x_n)\) converges to \(u\).
Suppose now that \((x_n)\) is bounded below and decreasing. Let \(w=\inf\{x_n\;|\; n\in\N\}\) and let \(\varepsilon \gt 0\) be arbitrary. Then by the properties of the infimum, there exists \(x_K\) such that \(w\leq x_K \lt w+\varepsilon\). Since \((x_n)\) is decreasing, and \(w\) is a lower bound for the range of the sequence, it follows that \(w\leq x_n\leq x_K\) for all \(n\geq K\). Therefore, \(w\leq x_n \lt w+\varepsilon\) for all \(n\geq K\). Hence, \(w-\varepsilon \lt x_n \lt w+\varepsilon\) for all \(n\geq K\). Since \(\varepsilon \gt 0\) was arbitrary, this proves that \((x_n)\) converges to \(w\).

We prove by induction that \(\frac{1}{2}\leq x_n\) for all \(n\in\N\), that is, \((x_n)\) is bounded below by \(\frac{1}{2}\). First of all, it is clear that \(\frac{1}{2}\leq x_1\). Now assume that \(\frac{1}{2}\leq x_n\) for some \(n\in\N\). Then \begin{align*} x_{n+1} &=\tfrac{1}{2}x_n + \frac{1}{4}\\ & \geq \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{4}\\ & = \frac{1}{2}. \end{align*} Hence, \((x_n)\) is bounded below by \(\tfrac{1}{2}\). We now prove that \((x_n)\) is decreasing. We compute that \(x_2 = \frac{1}{2}+\frac{1}{4} = \frac{3}{4}\), and thus \(x_2 \lt x_1\). Assume now that \(x_n \lt x_{n-1}\) for some \(n\in\N\). Then \[ x_{n+1} \lt \frac{1}{2} x_{n-1} + \frac{1}{4} = x_n. \] Hence, by induction we have shown that \((x_n)\) is decreasing. By the MCT, \((x_n)\) is convergent. Suppose that \((x_n)\rightarrow L\). Then also \((x_{n+1})\rightarrow L\) and the sequence \(y_n = \tfrac{1}{2}x_n + \frac{1}{4}\) converges to \(\tfrac{1}{2}L + \frac{1}{4}\). Therefore, \(L=\frac{1}{2} L + \frac{1}{4}\) and thus \(L = 1/2\).

We will prove by the MCT that \((x_n)\) converges. By induction, one can show that \(2^{n-1} \lt n!\) for all \(n\geq 3\). Therefore, \begin{align*} x_n & \lt 1 + \frac{1}{1}+\frac{1}{2} + \cdots + \frac{1}{2^{n-1}}\\ & = 1 + \frac{1-(1/2)^n}{1-(1/2)}\\ & \lt 1 + \frac{1}{1-(1/2)}\\ & = 3. \end{align*} Hence, \(x_n \lt 3\) and therefore \((x_n)\) is bounded. Now, since \(x_{n+1} = x_n + \frac{1}{(n+1)!}\) then clearly \(x_n \lt x_{n+1}\). Thus \((x_n)\) is increasing. By the MCT, \((x_n)\) converges. You might recognize that \(\limi x_n = e = 2.71828\ldots\).

Clearly, \(x_1 \lt x_2 = \sqrt{2}\). Assume by induction that \(x_k \gt x_{k-1}\) for some \(k\in\N\). Then \begin{align*} x_{k+1} &= \sqrt{2+x_k}\\ & \gt \sqrt{2+x_{k-1}}\\ & = x_k. \end{align*} Hence, \((x_n)\) is an increasing sequence. We now prove that \((x_n)\) is bounded above. Clearly, \(x_1 \lt 2\). Assume that \(x_k \lt 2\) for some \(k\in\N\). Then \(x_{k+1} = \sqrt{2+x_k} \lt \sqrt{2+2} = 2\). This proves that \((x_n)\) is bounded above. By the MCT, \((x_n)\) converges, say to \(L\). Moreover, since \(x_n \geq 0\) (as can be proved by induction), then \(L\geq 0\). Therefore, \(L = \sqrt{2+L}\) and then \(L^2 - L - 2 = 0\). Hence, \(L=\frac{1\pm\sqrt{1+8}}{2} = \frac{1\pm 3}{2}\). Since \(L\geq 0\) then \(L=2\).

Let \(\veps \gt 0\). Then there exists \(K\in\N\) such that \(|x_n-L| \lt \veps\) for all \(n\geq K\). Since \(n_K \geq K\) and \(n_{K} \lt n_{K+1} \lt \ldots\), then \(|x_{n_k}-L| \lt \veps\) for all \(k\geq K\).

Let \((x_n)\) be an arbitrary sequence. We will say that the term \(x_m\) is a peak if \(x_m\geq x_n\) for all \(n\geq m\). In other words, \(x_m\) is a peak if it is an upper bound of all the terms coming after it. There are two possible cases for \((x_n)\), either it has an infinite number of peaks or it has a finite number of peaks. Suppose that it has an infinite number of peaks, say \(x_{m_1}, x_{m_2},\ldots\), and we may assume that \(m_1 \lt m_2 \lt m_3 \lt \cdots\). Then, \(x_{m_1}\geq x_{m_2}\geq x_{m_3}\geq \cdots\), and therefore \((x_{m_k})\) is a decreasing sequence. Now suppose that there are only a finite number of peaks and that \(x_m\) is the last peak. Then \(x_{n_1}=x_{m+1}\) is not a peak and therefore there exists \(n_2 \gt n_1\) such that \(x_{n_2}\geq x_{n_1}\). Similarly, \(x_{n_2}\) is not a peak and therefore there exists \(n_3 \gt n_2\) such that \(x_{n_3}\geq x_{n_2}\). Hence, by induction, there exists a subsequence \((x_{n_k})\) that is increasing.

Let \((x_n)\) be an arbitrary bounded sequence. By Theorem 3.4.7, \((x_n)\) has a monotone subsequence \((x_{n_k})\). Since \((x_n)\) is bounded then so is \((x_{n_k})\). By the MCT applied to \((x_{n_k})\) we conclude that \((x_{n_k})\) is convergent.

If \((x_n)\) is a bounded sequence, then there exists \(a_1,b_1\in\real\) such that \(a_1\leq x_n\leq b_1\) for all \(n\in \N\). We will apply a recursive bisection algorithm to hunt down a converging subsequence of \((x_n)\). Let \(m_1 = \frac{(a_1+b_1)}{2}\) be the mid-point of the interval \([a_1,b_1]\). Then at least one of the subsets \(I_1=\{n\in\N\;:\; a_1\leq x_n \leq m_1\}\) or \(J_1=\{n\in\N\;:\; m_1\leq x_n \leq b_1\}\) is infinite; if it is \(I_1\) then choose some \(x_{n_1} \in [a_1,m_1]\) and let \(a_2=a_1\) and \(b_2= m_1\); otherwise choose some \(x_{n_1} \in [m_1,b_1]\) and let \(a_2=m_1\), \(b_2=b_1\). In any case, it is clear that \((b_2-a_2) = \frac{(b_1-a_1)}{2}\), that \(a_1\leq a_2\) and that \(b_1\geq b_2\). Now let \(m_2=\frac{(a_2+b_2)}{2}\) be the mid-point of the interval \([a_2,b_2]\) and let \(I_2=\{n\in\N\;:\; a_2\leq x_n \leq m_2\}\) and let \(J_2=\{n\in\N\;:\; m_2\leq x_n \leq b_2\}\). If \(I_2\) is infinite then choose some \(x_{n_2}\in [a_2,m_2]\) and let \(a_3=a_2\), \(b_3=m_2\); otherwise choose some \(x_{n_2}\in [m_2,b_2]\) and let \(a_3=m_2\) and \(b_3=b_2\). In any case, it is clear that \((b_3-a_3) = \frac{(b_1-a_1)}{2^2}\), that \(a_2\leq a_3\) and \(b_3\leq b_2\). By induction, there exists sequences \((a_k)\), \((b_k)\), and \((x_{n_k})\) such that \(a_k \leq x_{n_k}\leq b_k\) and \((b_k-a_k) = \frac{(b_1-a_1)}{2^{k-1}}\), \((a_k)\) is increasing and \((b_k)\) is decreasing. It is clear that \(a_k \leq b\) and \(a \leq b_k\) for all \(k\in \N\). Hence, by the MCT, \((a_k)\) and \((b_k)\) are convergent. Moreover, since \((b_k-a_k) = \frac{(b_1-a_1)}{2^{k-1}}\) then \(\lim (b_k-a_k) = 0\) and consequently \(\lim a_k = \lim b_k = L\). By the Squeeze theorem we conclude that \(\lim_{k\rightarrow\infty} x_{n_k} = L\).

Let \(u=\sup(S)\). If \(\veps \gt 0\) then there exists \(s\in S\) such that \(u-\veps \lt s \leq u\). Since \(s\in S\), there exists a subsequence \((x_{n_k})\) of \((x_n)\) that converges to \(s\). Therefore, there exists \(K\in\N\) such that \(u-\veps \lt x_{n_k} \lt u+\veps\) for all \(k\geq K\). Hence, for each \(\veps\), the inequality \(u-\veps \lt x_n \lt u+\veps\) holds for infinitely many \(n\). Consider \(\veps_1=1\). Then there exists \(x_{n_1}\) such that \(u-\veps_1 \lt x_{n_1} \lt u+\veps_1\). Now take \(\veps_2 =\frac{1}{2}\). Since \(u-\veps_2 \lt x_n \lt u+\veps\) holds for infinitely many \(n\), there exists \(x_{n_2}\) such that \(u-\veps_2 \lt x_{n_2} \lt u+\veps_2\) and \(n_2 \gt n_1\). By induction, for \(\eps_k = \frac{1}{k}\), there exists \(x_{n_k}\) such that \(u-\eps_k \lt x_{n_k} \lt u+\eps_k\) and \(n_k \gt n_{k-1}\). Hence, the subsequence \((x_{n_k})\) satisfies \(|x_{n_k} - u| \lt \frac{1}{k}\) and therefore \((x_{n_k})\rightarrow u\). Therefore, \(u = \sup(S) \in S\).

(i)\(\rightarrow\)(ii) Let \(S\) denote the subsequences limit set of \((x_n)\). By definition, \(L^*=\limsup x_n = \sup(S)\) and by Lemma 3.5.3 we have that \(L^* \in S\). Hence, there exists a subsequence of \((x_n)\) converging to \(L^*\) and thus \(L^*-\varepsilon \lt x_{n} \lt L^*+\varepsilon\) holds for infinitely many \(n\). In particular \(L^*-\varepsilon \lt x_{n}\) holds for infinitely many \(n\). Suppose that \(L^*+\varepsilon \lt x_n\) holds infinitely often. Now \(x_n\leq M\) for all \(n\) and some \(M \gt 0\). Since the inequality \(L^*+\varepsilon \lt x_n\) holds infinitely often, there exists a sequence \(n_1 \lt n_2 \lt \cdots\) such that \(L^*+\varepsilon \lt x_{n_k}\leq M\) for all \(k\in\N\). We can assume that \((x_{n_k})\) is convergent (because it is bounded and we can pass to a subsequence by the MCT) and thus \(\displaystyle L^*+\varepsilon\leq \limi x_{n_k}\leq M\). Hence we have proved that the subsequence \((x_{n_k})\) converges to a number greater than \(L^*\) which contradicts the definition of \(L^*=\sup(S)\).
(ii)\(\rightarrow\)(iii) Let \(\varepsilon \gt 0\). Since \(L^*+\varepsilon/2 \lt x_m\) holds for finitely many \(m\), there exists \(M\) such that \(x_m\leq L^*+\varepsilon/2\) for all \(m\geq M\). Hence, \(L^*+\varepsilon/2\) is an upper bound of \(\{x_n\;|\; n\geq m\}\) and thus \(u_m \lt L^*+\varepsilon\). Since \((u_m)\) is decreasing, we have that \(u_m \lt L^*+\varepsilon\) for all \(m\geq M\). Now, \(L^*-\varepsilon/2 \lt x_n\) holds infinitely often and thus \(L^*-\varepsilon \lt u_m\) for all \(m\in\N\). Hence, \(L^*-\varepsilon \lt u_m \lt L^*+\varepsilon\) for all \(m\geq M\). This proves the claim.
(iii)\(\rightarrow\)(i) Let \((x_{n_k})\) be a convergent subsequence. Since \(n_k\geq k\), by definition of \(u_k\), we have that \(x_{n_k}\leq u_k\). Therefore, \(\lim x_{n_k} \leq \lim u_k = L^*\). Hence, \(L^*\) is an upper bound of \(S\). By definition of \(u_k\), there exists \(x_{n_1}\) such that \(u_1-1 \lt x_{n_1}\leq u_1\). By induction, there exists a subsequence \((x_{n_k})\) such that \(u_k-\frac{1}{k} \lt x_{n_k}\leq u_k\). Hence, by the Squeeze Theorem, \(L^*=\lim x_{n_k}\). Hence, \(L^*\in S\) and thus \(L^*=\sup S=\liminf x_n\).

Suppose that \((x_n)\) does not converge to \(L\). Then, there exists \(\veps \gt 0\) such that for every \(K\in\N\) there exists \(n\geq K\) such that \(|x_n-L|\geq \veps\). Let \(K_1\in\mathbb{N}\). Then there exists \(n_1\geq K_1\) such that \(|x_{n_1}-L|\geq \veps\). Then there exists \(n_2 \gt n_1+1\) such that \(|x_{n_2}-L|\geq\veps\). By induction, there exists a subsequence \((x_{n_k})\) of \((x_n)\) such that \(|x_{n_k}-L|\geq \veps\) for all \(k\in\N\). Now \((x_{n_k})\) is bounded and therefore by Bolzano-Weierstrass has a convergent subsequence, say \((z_k)\), which is also a subsequence of \((x_n)\). By assumption, \((z_k)\) converges to \(L\), which contradicts that \(|x_{n_k}-L|\geq \veps\) for all \(k\in\N\).

Suppose that \((x_n)\rightarrow L\). Let \(\varepsilon \gt 0\) and let \(K\) be sufficiently large so that \(|x_n-L| \lt \varepsilon/2\) for all \(n\geq K\). If \(n,m\geq K\) then by the triangle inequality, \begin{align*} |x_n-x_m| &= |x_n-L+L-x_m|\\ &\leq |x_n-L| + |x_m-L|\\ & \lt \varepsilon/2 + \varepsilon/2\\ &= \varepsilon. \end{align*} This proves that \((x_n)\) is Cauchy.

The proof is similar to the proof that a convergent sequence is bounded.

In Lemma 3.6.3 we already showed that if \((x_n)\) converges then it is a Cauchy sequence. To prove the converse, suppose that \((x_n)\) is a Cauchy sequence. By Lemma 3.6.4, \((x_n)\) is bounded. Therefore, by the Bolzano-Weierstrass theorem there is a subsequence \((x_{n_k})\) of \((x_n)\) that converges, say it converges to \(L\). We will prove that \((x_n)\) also converges to \(L\). Let \(\eps \gt 0\) be arbitrary. Since \((x_n)\) is Cauchy there exists \(K\in\N\) such that if \(n,m\geq K\) then \(|x_n-x_m| \lt \eps/2\). On the other hand, since \((x_{n_k})\rightarrow L\) there exists \(n_M \geq K\) such that \(|x_{n_M}-L| \lt \eps/2\). Therefore, if \(n\geq K\) then \begin{align*} |x_n - L| &= |x_n - x_{n_M} + x_{n_M} - L|\\[2ex] &\leq |x_n - x_{n_M}| + |x_{n_M} - L| \\[2ex] & \lt \eps/2 + \eps/2\\ &= \eps. \end{align*} This proves that \((x_n)\) converges to \(L\).

Let \(S\subset\real\) be a non-empty set that is bounded above. If \(u\) is an upper bound of \(S\) and \(u\in S\) then \(u\) is the least upper bound of \(S\) and since \(u\in \real\) there is nothing to prove. Suppose then that no upper bound of \(S\) is an element of \(S\). Let \(a_1\in S\) be arbitrary and let \(b_1\in\real\) be an upper bound of \(S\). Then \(a_1 \lt b_1\), and we set \(M = b_1 - a_1 \gt 0\). Consider the mid-point \(m_1=\tfrac{a_1+b_1}{2}\) of the interval \([a_1,b_1]\). If \(m_1\) is an upper bound of \(S\) then set \(b_2 = m_1\) and set \(a_2 = a_1\), otherwise set \(b_2 = b_1\) and \(a_2 = m_1\). In any case, we have \(|a_2 - a_1| \leq \frac{M}{2}\), \(|b_2-b_1|\leq \frac{M}{2}\), and \(|b_2-a_2| = \frac{M}{2}\). Now consider the mid-point \(m_2 = \tfrac{a_2+b_2}{2}\) of the interval \([a_2,b_2]\). If \(m_2\) is an upper bound of \(S\) then set \(b_3 = m_2\) and \(a_3=a_2\), otherwise set \(b_3 = b_2\) and \(a_3 = m_2\). In any case, we have \(|a_3-a_2| \leq \frac{M}{2^2}\), \(|b_3-b_2| \leq \frac{M}{2^2}\), and \(|b_3-a_3| = \frac{M}{2^2}\). By induction, there exists a sequence \((a_n)\) such that \(a_n\) is not an upper bound of \(S\) and a sequence \((b_n)\) such that \(b_n\) is an upper bound of \(S\), and \(|a_n - a_{n+1}| \leq \frac{M}{2^n}\), \(|b_n - b_{n+1}| \leq \frac{M}{2^n}\), and \(|b_n - a_n| = \frac{M}{2^{n-1}}\). By Exercise 3.6.6, and using the fact that \(\limi r^n = 0\) if \(0 \lt r \lt 1\) (this is where the Archimedean property is needed), it follows that \((a_n)\) and \((b_n)\) are Cauchy sequences and therefore by assumption both \((a_n)\) and \((b_n)\) are convergent. Since \(|b_n - a_n| = \frac{M}{2^{n-1}}\) it follows that \(u = \lim a_n = \lim b_n\). We claim that \(u\) is the least upper bound of \(S\). First of all, for fixed \(x\in S\) we have that \(x \lt b_n\) for all \(n\in \N\) and therefore \(x\leq \lim b_n\), that is, \(u\) is an upper bound of \(S\). Since \(a_n\) is not an upper bound of \(S\), there exists \(x_n \in S\) such that \(a_n \lt x_n \lt b_n\) and therefore by the Squeeze theorem we have \(u = \lim x_n\). Given an arbitrary \(\eps \gt 0\) then there exists \(K\in \N\) such that \(u - \eps \lt x_K\) and thus \(u-\eps\) is not an upper bound of \(S\). This proves that \(u\) is the least upper bound of \(S\).

By definition, if \(\sum x_n\) converges then the sequence of partial sums \((s_n)\) is convergent. Suppose then that \(L = \limi s_n = \sum x_n\). Recall that \((s_n)\) has the recursive definition \(s_{n+1} = s_n + x_{n+1}\). The sequences \((s_{n+1})\) and \((s_n)\) both converge to \(L\) and thus \begin{align*} \limi x_{n+1} &= \limi (s_{n+1} - s_n)\\ & = L - L\\ & = 0. \end{align*}

Clearly, if \(\sum x_n = \limi s_n\) exists then \((s_n)\) is bounded. Now suppose that \((s_n)\) is bounded. Since \(x_n\geq 0\) for all \(n\in\N\) then \(s_{n+1} = s_n + x_{n+1} \geq s_n\) and thus \(s_{n+1}\geq s_n\) shows that \((s_n)\) is an increasing sequence. By the Monotone convergence theorem, \((s_n)\) converges and \(\limi s_n = \sup\{s_n\;|\; n\geq 1\}\).

Using the limit laws: \begin{align*} \limi \sum_{k=1}^n (x_k \pm y_k) &= \limi \left(\sum_{k=1}^n x_k \pm \sum_{k=1}^n y_k\right) \\[2ex] &= \limi \sum_{k=1}^n x_k \pm \limi \sum_{k=1}^n y_k\\[2ex] &= \sumo x_n \pm \sumo y_n. \end{align*} Hence, \[ \sumo (x_n \pm y_n) = \sumo x_n \pm \sumo y_n. \] If \(c\) is a constant then by the Limit Laws, \[ \limi \sum_{k=1}^n c x_k = c \limi \sum_{k=1}^n x_k = c \sumo x_n. \] Therefore, \[ \sumo c x_n = c \sumo x_n. \]

Let \(t_n = \sum_{k=1}^n x_n\) and \(s_n = \sum_{k=1}^n y_k\) be the sequences of partial sums. Since \(x_n\leq y_n\) then \(t_n \leq s_n\). To prove (a), if \(\sum y_n\) converges then \((s_n)\) is bounded. Thus, \((t_n)\) is also bounded. Since \((t_n)\) is increasing and bounded it is convergent by the MCT. To prove (b), if \(\sum x_n\) diverges then \((t_n)\) is necessarily unbounded and thus \((s_n)\) is also unbounded and therefore \((s_n)\) is divergent.

From \(-|x_n| \leq x_n \leq |x_n|\) we obtain that \[ 0 \leq x_n + |x_n| \leq 2|x_n|. \] If \(\sum |x_n|\) converges then so does \(\sum 2|x_n|\). By the Comparison test 3.7.18, the series \(\sum (x_n + |x_n|)\) converges also. Then the following series is a difference of two converging series and therefore converges: \[ \sum (x_n + |x_n|) - \sum |x_n| = \sum x_n \] and the proof is complete.

Suppose that \(L \lt 1\). Let \(\eps \gt 0\) be such that \(r = L+\eps \lt 1\). There exists \(K\in \N\) such that \(\frac{|x_{n+1}|}{|x_n|} \lt L+\eps = r\) for all \(n\geq K\) and thus \(|x_{n+1}| \lt |x_n| r\) for all \(n\geq K\). By induction, it follows that \(|x_{K+m}| \lt |x_K| r^m\) for all \(m\geq 1\). Since \(\sum_{m=1}^\infty |x_k| r^m\) is a geometric series with \(r \lt 1\) then, by the comparison test, the series \(\sum_{m=1}^\infty |x_{K+m}|\) converges. Therefore, the series \(\sum |x_n|\) converges and by the Absolute convergence criterion we conclude that \(\sum x_n\) converges absolutely. If \(L \gt 1\) then a similar argument shows that \(\sum x_n\) diverges. The case \(L=1\) follows from the fact that some series converge and some diverge when \(L=1\).