# Harrison Chapman

## Math 317: Homework 3

Due: Friday, February 15, 2019
1. (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 $% $. 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 $% $. Also $n > N$ implies $% $, so $\lim s_n = -\infty.$ $\Box$

2. 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 $% $ for all $n \in \mathbb N$. The base case is from that, $% $. Suppose now that $% $. Then,

Notice that this implies then that $% $. Hence $% $ 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. $\Box$.

3. (10.8) 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)

Proof. We will show that $\sigma_n \le \sigma_{n+1}$ by showing that $\sigma_{n+1} - \sigma_n$ is nonnegative.

As $n(n+1)$ is positive, this fraction is nonnegative if its numerator is. So, we examine:

% 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. $\Box$

4. For each of the following sequences, find the $\liminf$ and the $\limsup$.

1. $\lim a_n = 0$ so $\liminf a_n = \limsup a_n = 0$.

2. $\liminf b_n = -\infty$ and $\limsup b_n = +\infty$.

3. $\liminf c_n = -1$ and $\limsup c_n = 1$.

5. (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 :

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 $% $. Then there exists $N$ so that for all $n > N$, $s_n > -M$. But this means $% $. So $\lim s_n = -\infty$ for all $n > N$ as $M$ was arbitrary. The reverse direction follows by a similar argument. $\Box$