# 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. $a_n = \frac{(-1)^n}{n+1}$
2. $b_n = n \cos \left( \frac {n\pi}{5} \right)$
3. $c_n = (-1)^n - \frac 1n$
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

\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*}
5. 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.