# Harrison Chapman

## Math 317: Homework 4

Due: Friday, September 27, 2019

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 $\liminf$ and the $\limsup$.

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 $(s_n)$, $(t_n)$ 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 $$s \le t$$.

4. Prove that $\liminf s_n = -\limsup(-s_n)$ for any sequence $s_n$.

We have, using a result from homework 2, that

5. Let $(s_n)$ be an increasing sequence of positive numbers. Let $(\sigma_n)$ be defined by,

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.:

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.