Even as the infinite series is contained by the finite sum, and in the unbounded a bound is discerned, so the soul of immensity dwells in the minute, and in the narrowest limits there are no limits.

— Jakob Bernoulli, Tractatus de seriebus infinitis (1689)

Between 1689 and 1704, Jakob Bernoulli published in Basel five dissertations on infinite series, sums with infinitely many terms. It was a new and dangerous subject. Some of these sums give a number, like

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

Others grow without stopping. The hard part is knowing which kind each one is, because common sense is misleading.

The harmonic series

Consider the sum of the reciprocals of the whole numbers:

$$1+\frac12+\frac13+\frac14+\frac15+\cdots$$

Musicians called it harmonic, because $\frac12,\frac13,\frac14,\dots$ are the string lengths that give the harmonics of a note. The terms get smaller and smaller and tend to zero. Intuition says the sum must stop somewhere. But it grows too slowly to be fooled by:

Terms Approximate sum
10 2.93
1,000 7.49
1,000,000 14.39
$10^{43}$ 100.0

It takes more than $10^{43}$ terms to reach 100. Even so, Jakob showed that the sum passes any number.

Jakob’s proof

He split the series into blocks ending at squares. Take a number $a$ and the terms from $\frac1a$ to $\frac1{a^2}$. Setting $\frac1a$ aside, there are $a^2-a$ terms left, and all of them are at least $\frac1{a^2}$, the smallest. So

$$\frac1a+\left(\frac1{a+1}+\cdots+\frac1{a^2}\right)\geq\frac1a+\frac{a^2-a}{a^2}=\frac1a+1-\frac1a=1.$$

Each such block is worth at least 1. Now just chain them: from $\frac12$ to $\frac14$, from $\frac15$ to $\frac1{25}$, from $\frac1{26}$ to $\frac1{676}$, and so on:

$$1+\underbrace{\frac12+\frac13+\frac14}_{\geq1}+\underbrace{\frac15+\cdots+\frac1{25}}_{\geq1}+\underbrace{\frac1{26}+\cdots+\frac1{676}}_{\geq1}+\cdots$$

There are infinitely many blocks, each worth at least 1, so the sum passes any number. It diverges.

Jakob acknowledged that his brother Johann had found another proof before him, and published it in the same dissertation. Neither of them knew that the Frenchman Nicole Oresme had done the same around 1350, grouping the terms into blocks of size 1, 2, 4, 8, …

The squares: less than 2

And if the denominators are squares?

$$1+\frac14+\frac19+\frac1{16}+\frac1{25}+\cdots$$

Here Jakob showed the opposite: the sum does not pass 2. The trick is to compare each term with a difference. For $n\geq2$,

$$\frac1{n^2}<\frac1{n(n-1)}=\frac1{n-1}-\frac1n.$$

Adding from $n=2$ to $n=N$, almost everything cancels:

$$\begin{aligned}\sum_{n=2}^{N}\frac1{n^2}&<\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1{N-1}-\frac1N\right)\\&=1-\frac1N<1.\end{aligned}$$

Adding the first term, $1+\frac14+\frac19+\cdots<2$. The sum is a sequence that grows and never passes 2, so it reaches a number. Which number? Jakob could not find out.

The call for help

Jakob ended the discussion with a public appeal, in Latin, to the mathematicians of Europe:

If anyone finds and communicates to us what has so far eluded our efforts, we shall be very grateful.

— Jakob Bernoulli, Tractatus de seriebus infinitis (1689)

Nobody managed it during his lifetime. The question became known as the Basel problem, after the Bernoullis’ city. Johann tried too and failed. In 1735 a former pupil of Johann’s, Leonhard Euler, found the answer:

$$1+\frac14+\frac19+\frac1{16}+\cdots=\frac{\pi^2}6\approx1.6449.$$

When Johann heard of it, he lamented that his brother was not alive to see it.

Exercise: how many terms of the harmonic series are needed to pass 3?

The partial sums are $1$, $1.5$, $1.833$, $2.083$, $2.283$, $2.45$, $2.593$, $2.718$, $2.829$, $2.929$, $3.020$. It takes 11 terms: $1+\frac12+\cdots+\frac1{11}\approx3.02$.

Exercise: use the same idea to show that 1/1·2 + 1/2·3 + 1/3·4 + ... is exactly 1.

Since $\frac1{n(n+1)}=\frac1n-\frac1{n+1}$, the sum up to $N$ is

$$\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1N-\frac1{N+1}\right)=1-\frac1{N+1}.$$

As $N$ grows, $\frac1{N+1}$ goes to zero and the sum tends to $1$. Sums like this, where everything cancels except the ends, are called telescoping.

Sources: Jakob Bernoulli, Tractatus de seriebus infinitis (Basel, 1689–1704; collected at the end of the Ars conjectandi, 1713) · William Dunham, Journey through Genius (Wiley, 1990), ch. 8 · biography of Jacob Bernoulli, MacTutor.

All topics on James Bernoulli · ← Interest that never ends and the number e · The marvellous spiral and the lemniscate →

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