Reading: Read sections 10, 11, 12.
Exercises:
Come up with an example for each of the following, or explain why they cannot exist.
A decreasing, divergent sequence.
An unbounded, convergent sequence.
An increasing Cauchy sequence.
A sequence with undefined \(\liminf\).
A sequence whose subsequences all converge.
For instance \((s_n) = (-n)\)
Can’t exist; all convergent sequences are bounded.
Remember that any convergent sequence of reals is Cauchy. So for instance we can take \((s_n) = (-1/n)\) which is increasing and convergent (hence Cauchy).
Can’t exist; limit inferior is always defined in \(\mathbb R \cup \{\pm \infty\}\).
This is true of any convergent sequence, such as \((s_n) = (1)\)
For each of the following sequences, find the \(\liminf\) and the \(\limsup\).
\(\limsup a_n = 0\), \(\liminf a_n = 0\)
\(\limsup b_n = +\infty\), \(\liminf b_n = - \infty\)
\(\limsup c_n = 1\), \(\liminf c_n = -1\)
Let \((s_n)\), \((t_n)\) be sequences. Suppose there exists \(N_0\) so that \(s_n \le t_n\) for all \(n > N_0\). Prove that if both \((s_n), (t_n)\) converge, then \(\lim {s_n} \le \lim {t_n}\).
Let \(\lim s_n = s\) and \(\lim t_n = t\); we wish to show that \(L = \lim{(t_n - s_n)} = t - s \ge 0\). Suppose otherwise, that is, suppose \(L < 0\). Let \(\epsilon = -L/2 > 0\).
But now for any \(N\), consider \(n > \max\{N_0, N\}\): We see that \(\vert (t_n - s_n) - L \vert = (t_n - s_n) - L \ge -L > -L/2 = \epsilon\) (as both \(t_n - s_n\) and \(-L\) are positive when \(n > N_0\)). But this means that \(L < 0\) cannot be \(\lim{(t_n - s_n)}\). So we conclude \(s \le t\).
Prove that \(\liminf s_n = -\limsup(-s_n)\) for any sequence \(s_n\).
We have, using a result from homework 2, that
\[\begin{align*} \liminf s_n &= \lim_{N \to \infty} \sup\{ s_n : n > N \} \\ &= \lim_{N \to \infty} -\inf\{ -s_n : n > N \} \\ &= -\lim_{N \to \infty} \inf\{ -s_n : n > N \} \\ &= - \liminf (-s_n) \end{align*}\]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)
We consider \(\sigma_{n+1} - \sigma_n\); if this quantity is positive then \((\sigma_n)\) is increasing.:
\[\begin{align*} \sigma_{n+1} - \sigma_n &= \frac{s_1 + s_2 + \cdots + s_n + s_{n+1}}{n+1} - \frac{s_1 + s_2 + \cdots + s_n}{n} \\ &= \frac{n(s_1 + s_2 + \cdots + s_n + s_{n+1}) - (n+1)(s_1 + s_2 + \cdots + s_n)}{n(n+2)} \\ &= \frac{n s_{n+1} - s_1 - s_2 - \cdots - s_n}{n(n+2)} \\ &= \frac{(s_{n+1} - s_1) + (s_{n+1} -s_2) + \cdots + (s_{n+1} - s_n)}{n(n+2)}. \end{align*}\]As \(n(n+1) > 0\) and \(s_{n+1} - s_k \ge 0\) for all \(1 \le k \le {n}\) as \((s_n)\) is increasing, we get that this difference is positive, and so \((\sigma_n)\) is increasing.