(10.5) Complete the proof of Theorem 10.4 by showing that if \((s_n)\) is an unbounded decreasing sequence, that \(\lim s_n = -\infty\).
Proof. Let \((s_n)\) be an unbounded increasing sequence. Let \(M < 0\). Since the set \(\{s_n : n \in \mathbb N\}\) is unbounded and it is bounded above by \(s_1\), it must be unbounded below. Hence for some \(N \in \mathbb N\) we have \(s_N < M\). Also \(n > N\) implies \(s_n \le s_N < M\), so \(\lim s_n = -\infty.\) \(\Box\)
Let \((s_n)\) be a sequence defined by \(s_1 = \sqrt 2\) and \(s_{n+1} = \sqrt{2 + \sqrt {s_n}}\). Show that \((s_n)\) converges to a real number. You may use that if \(0 \le a \le b\) then \(0 \le \sqrt a \le \sqrt b\). (Hint: Use the Monotone Convergence Theorem.)
Proof. We first prove by induction that \(0 < s_n \le s_{n+1} \le 2\) for all \(n \in \mathbb N\). The base case is from that, \(0 < s_1 = \sqrt 2 \le \sqrt{ 2 + \sqrt{\sqrt 2} } = s_2 < \sqrt{4} = 2\). Suppose now that \(0 < s_k \le s_{k+1} \le 2\). Then,
\[\begin{align*} 0 < s_k &\le s_{k+1} \le 2 \\ \sqrt 0 < \sqrt {s_k} &\le \sqrt{s_{k+1}} \le \sqrt 2 \\ 2 + \sqrt 0 < 2 + \sqrt {s_k} &\le 2 + \sqrt{s_{k+1}} \le 2 + \sqrt 2 \\ \sqrt{2 + \sqrt 0} < \sqrt{2 + \sqrt {s_k}} &\le \sqrt{2 + \sqrt{s_{k+1}}} \le \sqrt{2 + \sqrt 2} \\ \sqrt{2} < s_{k+1} &\le s_{k+2} < \sqrt{2+\sqrt 2} \end{align*}\]Notice that this implies then that \(0 < s_{k+1} \le s_{k+2} < 2\). Hence \(0 < s_n \le s_{n+1} \le 2\) for all \(n \in \mathbb N\). This means that \((s_n)\) is monotone increasing and bounded, so the Monotone Convergence Theorem for sequences says that it must converge to some real number. \(\Box\).
(10.8) Let \((s_n)\) be an increasing sequence of positive numbers. Let \((\sigma_n)\) be defined by,
\[\sigma_n = \frac{s_1 + s_2 + \cdots + s_n}{n}.\]Prove that \((\sigma_n)\) is in increasing sequence. (The \(\sigma_n\) are called Cesaro means)
Proof. We will show that \(\sigma_n \le \sigma_{n+1}\) by showing that \(\sigma_{n+1} - \sigma_n\) is nonnegative.
\[\begin{align*} \sigma_{n+1} - \sigma_n &= \frac{s_1 + s_2 + \cdots + s_n + s_{n+1}}{n+1} + \frac{s_1 + s_2 + \cdots + s_n}{n} \\ &= \frac {n(s_1 + s_2 + \cdots + s_n + s_{n+1}) - (n+1)(s_1 + s_2 + \cdots + s_n)}{n(n+1)} \end{align*}\]As \(n(n+1)\) is positive, this fraction is nonnegative if its numerator is. So, we examine:
\(\begin{align*} n(s_1 + s_2 + \cdots + s_n + s_{n+1}) - (n+1)(s_1 + s_2 + \cdots + s_n) \\ &= n(s_1 + s_2 + \cdots + s_n) + ns_{n+1} - n(s_1 + s_2 + \cdots + s_n) - (s_1 + s_2 + \cdots + s_n) \\ &= ns_{n+1} - (s_1 + s_2 + \cdots + s_n) \\ &= (s_{n+1}-s_1) + (s_{n+1}-s_2) + \cdots + (s_{n+1}-s_n) \end{align*}\) As each of the \(s_{n+1}-s_i\) is nonnegative as \(s_{n+1} > s_i > 0\) for each \(1 \le i \le n\) (as \(s_n\) is positive and increasing) the whole sum is nonnegative. So the numerator is nonnegative, and we conclude that \(\sigma_{n+1} - \sigma_n\) is nonnegative, so \(\sigma_{n} \le \sigma_{n+1}\), so \((\sigma_n)\) is an increasing sequence. \(\Box\)
For each of the following sequences, find the \(\liminf\) and the \(\limsup\).
\(\lim a_n = 0\) so \(\liminf a_n = \limsup a_n = 0\).
\(\liminf b_n = -\infty\) and \(\limsup b_n = +\infty\).
\(\liminf c_n = -1\) and \(\limsup c_n = 1\).
(11.8) Prove that \(\liminf s_n = -\limsup(-s_n)\) for any sequence \(s_n\). You may use the result of exercise 5.4 without proving it.
Proof. We have that :
\[\begin{align*} \liminf s_n &= \lim_{N \to \infty} \inf \{ s_n : n > N \} \text{ by definition} \\ &= \lim_{N \to \infty} \left( - \sup \{ -s_n : n > N\} \right) \text{ by ex 5.4}\\ &= -\lim_{N \to \infty} \left( \sup \{ -s_n : n > N\} \right) \text{ by thm 9.2 and ex 9.10b}\\ &= -\limsup{ -s_n }. \end{align*}\]It remains to show the result of exercise 9.10(b) which says that \(\lim s_n = +\infty\) if and only if \(\lim -s_n = -\infty\). Suppose that \(\lim s_n = + \infty\). Suppose that \(M < 0\). Then there exists \(N\) so that for all \(n > N\), \(s_n > -M\). But this means \(-s_n < M\). So \(\lim s_n = -\infty\) for all \(n > N\) as \(M\) was arbitrary. The reverse direction follows by a similar argument. \(\Box\)