Department of Mathematics

Colorado State University

Math 317: Homework 4

Due: Friday, September 27, 2019

Reading: Read sections 10, 11, 12.

Exercises:

  1. Come up with an example for each of the following, or explain why they cannot exist.

    1. A decreasing, divergent sequence.

    2. An unbounded, convergent sequence.

    3. An increasing Cauchy sequence.

    4. A sequence with undefined \(\liminf\).

    5. A sequence whose subsequences all converge.

    1. For instance \((s_n) = (-n)\)

    2. Can’t exist; all convergent sequences are bounded.

    3. 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).

    4. Can’t exist; limit inferior is always defined in \(\mathbb R \cup \{\pm \infty\}\).

    5. This is true of any convergent sequence, such as \((s_n) = (1)\)

  2. For each of the following sequences, find the and the .

    1. \(\limsup a_n = 0\), \(\liminf a_n = 0\)

    2. \(\limsup b_n = +\infty\), \(\liminf b_n = - \infty\)

    3. \(\limsup c_n = 1\), \(\liminf c_n = -1\)

  3. Let , 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 \(s \le t\).

  4. Prove that for any sequence .

    We have, using a result from homework 2, that

  5. Let be an increasing sequence of positive numbers. Let be defined by,

    Prove that is in increasing sequence. (The are called Cesaro means)

    We consider \(\sigma_{n+1} - \sigma_n\); if this quantity is positive then \((\sigma_n)\) is increasing.:

    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.