Write

$$1+\frac12+\frac14+\frac18+\cdots=2.$$

It looks like a calculation such as $3+4=7$: terms on the left, result on the right. But nobody has ever added up all these terms, nor ever will. The equals sign here does a job different from all those chapter 4 listed. It states that the partial sums

$$1,\quad1.5,\quad1.75,\quad1.875,\quad\ldots$$

get as close to $2$ as you like. In symbols,

$$\sum_{n=0}^{\infty}\frac1{2^n}=\lim_{N\to\infty}\sum_{n=0}^{N}\frac1{2^n}=2.$$

The $\infty$ on top of the $\Sigma$ is the same direction-$\infty$ as in the previous text: it is not a last term, it is the instruction to keep going.

0.999… = 1

The equality that unsettles students most is

$$0.999\ldots=1.$$

Many think that $0.999\ldots$ “gets very close, but never arrives” at $1$. The objection has a semiotic origin: the dots seem to describe a process that is going on, and a process does not end. But $0.999\ldots$ is not a process; it is the name of the limit of that process:

$$0.999\ldots=\frac9{10}+\frac9{100}+\frac9{1000}+\cdots=\frac{9/10}{1-1/10}=1.$$

The researchers David Tall and Shlomo Vinner, in a 1981 study that became a classic, called this a conflict between the concept image, what the student pictures on seeing the symbol, and the concept definition. The image is “something that gets closer”; the definition is “the number that everything gets closer to”.

Sums that deceive

When the partial sums do not approach anything, the writing with $=$ stays the same, and that is where the danger lies.

Grandi’s series. In 1703 the Italian priest Guido Grandi considered

$$1-1+1-1+1-\cdots$$

Grouping one way, $(1-1)+(1-1)+\cdots=0$. Another way, $1-(1-1)-(1-1)-\cdots=1$. Grandi concluded that the sum was $\frac12$, and saw in it an image of the creation of the world out of nothing. The partial sums, $1,0,1,0,\ldots$, do not approach anything: the series has no sum in the sense of the limit.

Euler’s $-1$. The formula for the geometric series, $1+x+x^2+\cdots=\frac1{1-x}$, holds for $|x|<1$. Euler applied it with $x=2$ and wrote

$$1+2+4+8+\cdots=\frac1{1-2}=-1.$$

A sum of positive numbers coming out negative. The notation accepted the substitution without protest; it took the mathematicians of the nineteenth century to demand that every $=$ with dots come with a proof of convergence.

Order matters

In finite sums, the order of the terms does not matter: $3+4=4+3$. In infinite ones, it can. The alternating harmonic series

$$1-\frac12+\frac13-\frac14+\frac15-\cdots=\ln2\approx0.693$$

converges. Now rearrange the same terms, putting two negatives after each positive:

$$1-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac1{10}-\frac1{12}+\cdots$$

Grouping in threes, each group is $\frac1{2k-1}-\frac1{2(2k-1)}-\frac1{4k}=\frac12\left(\frac1{2k-1}-\frac1{2k}\right)$, and the whole sum becomes half of the original: $\frac12\ln2$. Bernhard Riemann proved that, for series like this, a suitable rearrangement gives any value you like. The sign $\Sigma$ does not show the order of the terms, and here order is everything.

Exercise: 0.333… and 0.999…

Write $0.333\ldots$ as a geometric series and calculate the sum. Then use the result to obtain $0.999\ldots$

$0.333\ldots=\frac3{10}+\frac3{100}+\cdots=\frac{3/10}{1-1/10}=\frac13$. Multiplying by 3: $0.999\ldots=3\cdot\frac13=1$. Whoever accepts $0.333\ldots=\frac13$ and rejects $0.999\ldots=1$ is rejecting a multiplication by 3.

Exercise: where Achilles catches up

With the data from the curiosity box, find the meeting point without a series, and check it with the series.

Let $d$ be the distance Achilles runs until he catches the tortoise. It runs $\frac{d}{10}$ in the same time, and $d=100+\frac d{10}$, so $d=\frac{1000}9=111.1\ldots$ metres. The geometric series gives the same: $\frac{100}{1-1/10}=\frac{1000}{9}$.

Chapter index · ← Wallis’s ∞: a sign that is not a number · How to read calculus: a guide and a glossary →