# Thoughts on Infinity: Index

I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students.

Part 1: Different types of countable sets

Part 2a: Divergence of the harmonic series.

Part 2b: Convergence of the Kempner series.

Part 3a: Conditional convergent series or products shouldn’t be rearranged.

Part 3b: Definition of the Euler-Mascheroni constant $\gamma$.

Part 3c: Evaluation of the conditionally convergent series $\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} \dots$

Part 3d: Confirmation of this evaluation using technology.

Part 3e: Evaluation of a rearrangement of this conditionally convergent series.

Part 3f: Confirmation of this different evaluation using technology.

Part 3g: Closing thoughts.

# Error involving countable numbers in Glencoe Algebra 2 (2014)

Errors in textbooks happened when Pebbles Flintstone and Bamm-Bamm Rubble attended Flintstone Elementary, and they still happen on occasion today. But even with that historical perspective, this howler is a doozy.

This was sent to me by a former student of mine. It appears in the chapter study guide for Section 2.1 of Glencoe’s Algebra 2 textbook (published in 2014), presumably as an enrichment activity for students learning about the definitions of “one to one functions” and “onto functions.” Just how bad is this mistake?

• The above “proof” is only a blatant assertion, without any justification, either formal or informal, for why the authors think that the statement is false.
• The ordering of the rational numbers in the way listed above is actually reasonably close to the listing that actually does produce the one-to-one correspondence between $\mathbb{Q}$ and $\mathbb{Z}$.
• Just above Example 2 was Example 1, which gives the correct proof that there’s a one-to-one correspondence between $\mathbb{Z}$ and $\mathbb{N}$. If the authors had double-checked this proof in any reputable book, they should have also been able to double-check that their Example 2 was completely false.  The full chapter study guide can be found here (it’s on the last page): http://nseuntj.weebly.com/uploads/1/8/2/0/18201983/2.1relations_and_functions.pdf

Reactions can be found here: https://www.reddit.com/r/math/comments/3k1qe6/this_is_in_a_high_school_math_textbook_in_texas/

Reference to this can be seen on page 10 of the teacher’s manual here: http://msastete.com/yahoo_site_admin1/assets/docs/Chpte2-1.25882808.pdf

# Thoughts on Infinity (Part 3a)

Last summer, Math With Bad Drawings had a nice series on the notion of infinity that I recommend highly. This topic is a perennial struggle for math majors to grasp, and I like the approach that the author uses to sell this difficult notion.

Part 3 on infinite series and products that are conditionally convergent discusses a head-scratching fact: according to the Riemann series theorem, the commutative and associative laws do not apply to conditionally convergent series.

An infinite series $\displaystyle \sum_{n=1}^\infty a_n$ converges conditionally if it converges to a finite number but $\displaystyle \sum_{n=1}^\infty |a_n|$ diverges. Indeed, by suitably rearranging the terms, the sum can be changed so that the (rearranged) series converges to any finite value. Even worse, the terms can be rearranged so that the sum converges to either $\infty$ or $-\infty$. (Of course, this can’t happen for finite sums, and rearrangements of an absolutely convergent series do not change the value of the sum.)

I really like Math With Bad Drawing’s treatment of the subject, as it starts with an infinite product for $\pi/2$: The top line is correct. However, the bottom line has to be incorrect since $\pi/2 > 1$ but each factor on the right-hand side is less than 1. The error, of course, stems from conditional convergence (the terms in the top product cannot be rearranged).

Conditional convergence is typically taught but glossed over in Calculus II since these rearrangements are such a head-scratching topic. I really like the above example because the flaw in the logic is made evidence after only three steps.

In tomorrow’s post, I’ll continue with another example of rearranging the terms in a conditionally convergent series.

# Thoughts on Infinity (Part 2b)

Last summer, Math With Bad Drawings had a nice series on the notion of infinity that I recommend highly. This topic is a perennial struggle for math majors to grasp, and I like the approach that the author uses to sell this difficult notion.

Here’s Part 2 on the harmonic series, which is extremely well-written and which I recommend highly. Here’s a brief summary: the infinite harmonic series $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$

diverges. However, if you eliminate from the harmonic series all of the fractions whose denominator contains a 9, then the new series converges! This series has been called the Kempner series, named after the mathematician who first published this result about 100 years ago. To prove this, we’ll examine the series whose denominators are between 1 and 8, between 10 and 88, between 100 and 888, etc. First, each of the terms in the partial sum $\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}$

is less than or equal to $1$, and so the sum of the above eight terms must be less than $8$.

Next, each of the terms in the sum $\displaystyle \frac{1}{10} + \frac{1}{11} + \dots + \frac{1}{88}$

is less than $\displaystyle \frac{1}{10}$. Notice that there are $72$ terms in this sum since there are 8 possibilities for the first digit of the denominator (1 through 8) and 9 possibilities for the second digit (0 through 8). So the sum of these 72 terms must be less than $\displaystyle 8 \times \frac{9}{10}$.

Next, each of the terms in the sum $\displaystyle \frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{888}$

is less than $\displaystyle \frac{1}{100}$. Notice that there are $8 \times 9 \times 9$ terms in this sum since there are 8 possibilities for the first digit of the denominator (1 through 8) and 9 possibilities for the second and third digits (0 through 8). So the sum of these $8 \times 9 \times 9$ terms must be less than $8 \times \displaystyle \frac{9^2}{100}$.

Continuing, we see that the Kempner series is bounded above by $\displaystyle 8 + 8 \times \frac{9}{10} + 8 \times \frac{9^2}{10^2} + \dots$

Using the formula for an infinite geometric series, we see that the Kempner series converges, and the sum of the Kempner series must be less than $8 \times \displaystyle \frac{1}{1-9/10} = 80$.

Using the same type of reasoning, much sharper bounds for the sum of the Kempner series can also be found. This 100-year-old article from the American Mathematical Monthly demonstrates that the sum of the Kempner series is between $22.4$ and $23.3$.  For more information about approximating the sum of the Kempner series, see Mathworld and Wikipedia.

It should be noted that there’s nothing particularly special about the number $9$ in the above discussion. If all denominators containing $314159265$, or any finite pattern, are eliminated from the harmonic series, then the resulting series will always converge.

# Thoughts on Infinity (Part 2a)

Last summer, Math With Bad Drawings had a nice series on the notion of infinity that I recommend highly. This topic is a perennial struggle for math majors to grasp, and I like the approach that the author uses to sell this difficult notion.

Here’s Part 2 on the harmonic series, which is extremely well-written and which I recommend highly. Here’s a brief summary: the infinite harmonic series $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$

diverges. This is a perennial head-scratcher for students, as the terms become smaller and smaller yet the infinite series diverges.

To show this, notice that $\displaystyle \frac{1}{3} + \frac{1}{4} > \displaystyle \frac{1}{4} + \frac{1}{4} = \displaystyle \frac{1}{2}$, $\displaystyle \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \displaystyle \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \displaystyle \frac{1}{2}$,

and so on. Therefore, $\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} > \displaystyle 1 + \frac{1}{2} + \frac{1}{2} = 2$, $\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} +\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \displaystyle 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \displaystyle \frac{5}{2}$,

and, in general, $\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2^n} > \displaystyle 1+\frac{n}{2}$.

Since $\displaystyle \lim_{n \to \infty} \left(1 + \frac{n}{2} \right) = \infty$, we can conclude that the harmonic series diverges.

However, here’s an amazing fact which I hadn’t known before the Math With Bad Drawings post: if you eliminate from the harmonic series all of the fractions whose denominator contains a 9, then the new series converges!

I’ll discuss the proof of this fact in tomorrow’s post. Until then, here’s a copy of the comic used in the Math With Bad Drawings post. # Thoughts on Infinity (Part 1)

Last summer, Math With Bad Drawings had a nice series on the notion of infinity that I recommend highly. This topic is a perennial struggle for math majors to grasp, and I like the approach that the author uses to sell this difficult notion.