Exercise 7.3.2

July 30, 2020 Issa Rice Leave a comment

Exercise statement

Prove Lemma 7.3.3.

Lemma 7.3.3 (Geometric series). Let $x$ be a real number. If $|x| \geq 1$ , then the series $\sum_{n=0}^\infty x^n$ is divergent. If however $|x| < 1$ , then the series is absolutely convergent and

$\displaystyle \sum_{n=0}^\infty x^n = 1/(1-x)$ .

Hints

For the first part, use the zero test.
For the second part, first use induction to establish the geometric series formula $\sum_{n=0}^N x^n = (1-x^{N+1})/(1-x)$ and then apply Lemma 6.5.2.

How to think about the exercise

In the case where $|x| < 1$ , Tao suggests to “use induction”. I think this is a pretty unhelpful hint. It nudges you toward a particular unenlightening kind of induction proof where you just blindly go through the motions – it is a totally valid proof, and will work without any trouble, but I just don’t like it. Here I want to walk through a better proof.

Let $N$ be a natural number. The $N$ th partial sum of the series is

$\displaystyle S_N := \sum_{n=0}^N x^n = 1 + x + x^2 + \cdots + x^{N-1} + x^N$

The big insight is to notice that if you multiply this by $x$ , you almost get the $(N+1)$ th partial sum, $S_{N+1}$ . In fact, we have

$\displaystyle 1 + xS_N = 1 + x + x^2 + x^3 + \cdots + x^N + x^{N+1} = S_{N+1}$

But for any series, the difference between the $(N+1)$ th partial sum and the $N$ th partial sum is just the $(N+1)$ th term. In other words, we have $S_{N+1} - S_N = x^{N+1}$ . Substituting the identity $S_{N+1} = 1 + xS_N$ from above, we have

$x^{N+1} = S_{N+1} - S_N = (1 + xS_N) - S_N = 1 + (x-1)S_N$

Rearranging, we obtain the desired formula

$\displaystyle S_N = \frac{1 - x^{N+1}}{1-x}$

Dividing by $1-x$ is valid since $|x| < 1$ hence $x\ne 1$ .

The above wasn’t entirely rigorous because we used a bunch of “ $\cdots$ ” to make it easier to see the pattern. But it is very simple to convert this into a rigorous proof, which we will do below in the model solution.

Model solution

First suppose $|x| \geq 1$ . Consider the terms of this series $1, x, x^2, x^3, \ldots$ . By Lemma 6.5.2, this sequence either converges to $1$ or else diverges. In either case, the sequence does not converge to zero, so by the zero test (Corollary 7.2.6) the series $\sum_{n=0}^\infty x^n$ diverges.

Now suppose $|x| < 1$ . We will show the geometric series formula $\sum_{n=0}^N x^n = (1-x^{N+1})/(1-x)$ . Write $S_N$ for the $N$ th partial sum. Then we have

$\displaystyle \begin{aligned}1 + xS_N &= 1 + x\sum_{n=0}^N x^n \\ &= 1 + \sum_{n=0}^N x^{n+1} \\ &= 1 + \sum_{n=1}^{N+1} x^n \\ &= \sum_{n=0}^{N+1} x^n \\ &= S_{N+1}\end{aligned}$

We also have $S_{N+1} - S_N = x^{N+1}$ for any series. Combining these two equalities and rearranging, we have $S_N = (1-x^{N+1})/(1-x)$ as desired.

To show that the series $\sum_{n=0}^\infty x^n$ is convergent, we directly compute the limit $\lim_{N\to\infty} S_N$ . By Lemma 6.5.2 we have $\lim_{N\to\infty} x^{N+1} = 0$ . Thus by the limit laws we have

$\displaystyle \lim_{N\to\infty} S_N = \lim_{N\to\infty} \frac{1 - x^{N+1}}{1-x} = \frac{1 - \lim_{N\to\infty} x^{N+1}}{1-x} = \frac{1}{1-x}$

as desired.

Finally we show that when $|x| < 1$ the series is absolutely convergent. We must show that the series $\sum_{n=0}^\infty |x^n|$ converges. By Proposition 4.3.10(d) (which we can use thanks to Proposition 5.6.3), we have $|x^n| = |x|^n$ . Thus it suffices to show that the series $\sum_{n=0}^\infty |x|^n$ converges. But this is just a geometric series with the common ratio $|x|$ , so the series converges (to $\sum_{n=0}^\infty |x|^n = 1/(1-|x|)$ ) by our proof above.

Solutions

Exercise 7.2.4

July 29, 2020 Issa Rice 5 Comments

Exercise statement

Prove Proposition 7.2.9.

Proposition 7.2.9 (Absolute convergence test). Let $\sum_{n=m}^\infty a_n$ be a formal series of real numbers. If this series is absolutely convergent, then it is also conditionally convergent. Furthermore, in this case we have the triangle inequality

$\displaystyle \left|\sum_{n=m}^\infty a_n\right| \leq \sum_{n=m}^\infty |a_n|$

Hints

Use Proposition 7.2.5.
Use Proposition 7.1.4(e).

How to think about the exercise

In the proof below, we use Proposition 7.2.5 twice: once to go from the fact that $\sum_{n=m}^\infty a_n$ absolutely converges to an inequality, and a second time to go from an inequality to the fact that $\sum_{n=m}^\infty a_n$ conditionally converges. Make sure you introduce $\varepsilon > 0$ at the start of the proof, before you use Proposition 7.2.5 for the first time.

Model solution

Suppose $\sum_{n=m}^\infty a_n$ is absolutely convergent. We want to show this series is conditionally convergent. To do so, we will use Proposition 7.2.5, which is also known as Cauchy’s convergence test. So let $\varepsilon > 0$ be a real number. Since $\sum_{n=m}^\infty a_n$ is absolutely convergent, this means that $\sum_{n=m}^\infty |a_n|$ is a convergent series. Thus there exists an integer $N \geq m$ such that $\left|\sum_{n=p}^q |a_n|\right| \leq \varepsilon$ for all $p,q \geq N$ . The quantity $\sum_{n=p}^q |a_n|$ is non-negative for all $p,q$ (this is an easy induction), so in fact we have $\left|\sum_{n=p}^q |a_n|\right| = \sum_{n=p}^q |a_n|$ which means that $\sum_{n=p}^q |a_n| \leq \varepsilon$ for all $p,q \geq N$ .

By the triangle inequality of Proposition 7.1.4(e), we have $\left|\sum_{n=p}^q a_n\right| \leq \sum_{n=p}^q |a_n|$ for all $p,q$ . Thus we have found an $N\geq m$ such that for all $p,q \geq N$ we have

$\displaystyle \left|\sum_{n=p}^q a_n\right| \leq \sum_{n=p}^q |a_n| \leq \varepsilon$

which proves convergence of $\sum_{n=m}^\infty a_n$ .

It remains to show the triangle inequality, $\left|\sum_{n=m}^\infty a_n\right| \leq \sum_{n=m}^\infty |a_n|$ . Let $\alpha_N := \left|\sum_{n=m}^N a_n\right|$ be the absolute value of the $N$ th partial sum of the series $\sum_{n=m}^\infty a_n$ , and let $\beta_N := \sum_{n=m}^N |a_n|$ be the $N$ th partial sum of the series $\sum_{n=m}^\infty |a_n|$ . By Proposition 7.1.4(e), we have $\alpha_N \leq \beta_N$ for each $N\geq m$ . Thus by the comparison principle, Lemma 6.4.13, we have the inequality $\limsup_{N\to\infty} \alpha_N \leq \limsup_{N\to\infty} \beta_N$ .

We proved above that the series $\sum_{n=m}^\infty a_n$ converges. Thus by the limit laws, Theorem 6.1.19 (c) and (g) (and the fact that $|x| = \max(x, -x)$ for all real numbers $x$ ), we know that the sequence $(\alpha_N)_{N=m}^\infty$ converges to $\left|\sum_{n=m}^\infty a_n\right|$ (i.e. we know that the series converges, so this means the sequence of partial sums converges, which means by the limit laws that the sequence of absolute values of the partial sums also converges). Thus by Proposition 6.4.12(f) we have $\limsup_{N\to\infty} \alpha_N = \lim_{N\to\infty} \alpha_N$ .

Similarly, by the hypothesis that $\sum_{n=m}^\infty a_n$ is absolutely convergent, we know that the sequence $(\beta_N)_{N=m}^\infty$ converges. Thus $\limsup_{N\to\infty} \beta_N = \lim_{N\to\infty} \beta_N$ .

Combining these equalities with the inequality from above, we have

$\displaystyle \left|\sum_{n=m}^\infty a_n\right| = \lim_{N\to\infty} \alpha_N \leq \lim_{N\to\infty} \beta_N = \sum_{n=m}^\infty |a_n|$

as desired.

Solutions

Exercise 3.2.1

July 1, 2020 Issa Rice 5 Comments

Exercise statement

Show that the universal specification axiom, Axiom 3.8, if assumed to be true, would imply Axioms 3.2, 3.3, 3.4, 3.5, and 3.6. (If we assume that all natural numbers are objects, we also obtain Axiom 3.7.) Thus, this axiom, if permitted, would simplify the foundations of set theory tremendously (and can be viewed as one basis for an intuitive model of set theory known as “naive set theory”). Unfortunately, as we have seen, Axiom 3.8 is “too good to be true”!

Hints

None.

How to think about the exercise

This is a pretty straightforward exercise, so I don’t have anything to say.

Model solution

Axiom 3.2: Let $P(x)$ be any statement pertaining to an object $x$ that is always false, for instance the statement $x \ne x$ or the statement $0=1$ . (You might not like using $x\ne x$ because equality might not be defined for all objects, but typically in formal systems equality is defined for all objects and has the properties listed in Appendix A.7.) Then by Axiom 3.8 the set $\{x : P(x)\text{ is true}\}$ exists. In other words, we have $y \in \{x : P(x)\text{ is true}\}$ if and only if $P(y)$ is true. Since $P(y)$ is always false, this means that $y \in \{x : P(x)\text{ is true}\}$ is also always false, which means the set does not contain any elements, i.e. for every object $y$ we have $y \notin \{x : P(x)\text{ is true}\}$ . This is exactly what Axiom 3.2 claims, so we are done.

Axiom 3.3: Let $a$ be an object, and consider the statement $P(x)$ defined as $x=a$ . By Axiom 3.8, the set $\{x : x=a\}$ exists. Thus the singleton set $\{a\}$ exists. If $a,b$ are objects, we can consider the statement $Q(x)$ defined as “ $x=a$ or $x=b$ ”. Then by Axiom 3.8 the set $\{x : x=a\text{ or }x=b\}$ exists, which is the pair set $\{a,b\}$ . This proves Axiom 3.3.

Axiom 3.4: Let $A,B$ be two sets. Consider the statement $P(x)$ defined as “ $x\in A$ or $x\in B$ ”. Then by Axiom 3.8 the set $\{x : P(x)\text{ is true}\}$ exists. Thus we have $y \in \{x : P(x)\text{ is true}\} \iff (y \in A \text{ or }y \in B)$ , which means $\{x : P(x)\text{ is true}\}$ is the union of $A$ and $B$ .

Axiom 3.5: Let $A$ be a set, and let $P(x)$ be a statement pertaining to an object $x \in A$ . Let $Q(x)$ be the statement “ $x \in A$ and $P(x)$ is true”. Then by Axiom 3.8 the set $\{x : Q(x)\text{ is true}\}$ exists. Thus we have $y \in \{x : Q(x)\text{ is true}\} \iff (y \in A \text{ and }P(y)\text{ is true})$ . This is what we mean by the set $\{x \in A : P(x)\}$ , so this proves Axiom 3.5.

Axiom 3.6: Let $A$ be a set, and let $P(x,y)$ be a statement pertaining to $x,y$ such that for each $x \in A$ there is at most one $y$ for which $P(x,y)$ is true. Let $Q(y)$ be the statement “ $P(x,y)$ is true for some $x \in A$ ”. Then by Axiom 3.8, the set $\{y : Q(y) \text{ is true}\}$ exists. Thus we have $z \in \{y : Q(y) \text{ is true}\}$ iff $Q(z)$ is true iff $P(x,z)$ is true for some $x \in A$ . This is precisely what is meant by the set $\{y : P(x,y) \text{ is true for some } x \in A\}$ , so this proves Axiom 3.6.

Axiom 3.7: Suppose all natural numbers are objects. Let $P(x)$ be the statement “ $x$ is a natural number”. Then the set $\{x : P(x)\text{ is true}\}$ exists by Axiom 3.8. This set is $\mathbf N$ .