The Basel problem — what it asks, and how it was solved

Some problems are hard because they are complicated. The Basel problem is hard for the opposite reason: it takes one line to state, anyone can check the first few terms by hand, and yet for almost ninety years nobody could finish it. The question, which Pietro Mengoli asked in 1650 [1], is the value of the sum

the sum of the reciprocals of the squares: one, plus a quarter, plus a ninth, plus a sixteenth, and so on without end.

It was a natural question to ask. Nicole Oresme had shown in the fourteenth century that the harmonic series $1 + \frac12 + \frac13 + \cdots$ grows without bound, and Mengoli had just proved that the alternating series $1 - \frac12 + \frac13 - \cdots$ adds up to $\ln 2$. Squaring the denominators makes the terms shrink much faster, so this sum stays finite; the question is what it equals. Jakob Bernoulli proved that it is less than 2, spread the question, and admitted that its exact value had defeated him [2].¹ The problem is named after Basel, the city of the Bernoullis and of the man who finally solved it, Leonhard Euler, in 1735 [3]. Ayoub tells the whole story [4].

This article tells the story in the order it happened: the problem first, and why it was so stubborn, then the proofs one after another, from Euler’s to a double-integral proof found in 1983 and given its present form in 1993. Each of them finds the $\pi$ somewhere else.

Nobody can add infinitely many numbers one after another; the additions would never end. What we can do is add the first few and watch what happens. The sum of the first $N$ terms is called the $N$-th partial sum,

and the infinite sum is defined as the number these partial sums approach as $N$ grows, if there is one.

A simpler series shows the idea. Walk halfway across a room, then half the remaining distance, then half of what is left, and so on. The distances are $\frac12, \frac14, \frac18, \dots$, and after $N$ steps you have covered $1 - 1/2^N$ of the room. You never arrive, but the gap $1/2^N$ shrinks below any bound you care to name, and so

That is all the equals sign means for an infinite sum.

The Basel problem therefore asks two questions in one. Do the partial sums of (1) approach a limit at all? And if they do, which number is it?

For a series to converge, its terms must shrink to zero; otherwise every step adds at least some fixed amount, and the partial sums run off. But shrinking is not enough, and the harmonic series $1 + \frac12 + \frac13 + \frac14 + \cdots$ is the standard warning. Its terms shrink to zero, yet its partial sums grow without bound. The argument, which goes back to Oresme, groups the terms in blocks that end at powers of 2:

Each block adds up to at least $\frac12$. The two terms in the first bracket are each at least $\frac14$; the four in the second are each at least $\frac18$; the eight in the next are each at least $\frac{1}{16}$, and so on. So the first $2^k$ terms add up to at least $1 + k/2$, which passes any number you name once $k$ is large enough. The growth is extremely slow, though: the partial sums pass 5 at the 83rd term, and 10 only at the 12,367th. A series can diverge while looking, to anyone adding terms by hand, as if it were settling down.

Signs can change the picture. Mengoli’s alternating series $1 - \frac12 + \frac13 - \frac14 + \cdots$ has the same terms, but each one partly cancels the one before, and the partial sums settle down to $\ln 2 = 0.693\ldots$. The Basel series has no such help: all its terms are positive, so the only question is whether they shrink fast enough.

They do. The squares grow much faster than the whole numbers, so their reciprocals shrink much faster: the thousandth term of the harmonic series is one thousandth, while the thousandth term of (1) is one millionth. Jakob Bernoulli turned that into a proof by comparing the series with one whose partial sums can be computed exactly.

The telescoping is easiest to see written out. For $N = 4$ the comparison sum is

because each fraction subtracted in one bracket is added back in the next, and the sum collapses like the sections of a telescope sliding into one another. However many terms we take, only the first and the last survive.

Then there is the step from bounded to convergent, which is subtler than it looks. The partial sums only go up, since every term is positive, and they never pass 2. A sequence that keeps increasing but stays below a ceiling cannot wander: it has to crowd towards some number at or below the ceiling. That such a sequence always has a limit is a basic property of the real numbers, called completeness, and it is why the proof can promise a sum without saying what the sum is.

So the sum lies somewhere between 1 and 2, and the proof says nothing about where. That is the hard part, and it is the part Euler solved. The answer, which Sections 4 to 7 prove in four different ways, is this.

It is a strange answer. The number $\pi$ comes from circles: it is the ratio of a circle’s circumference to its diameter. Nothing in (1) mentions a circle, and the sum is built from whole numbers alone.

There is a second oddity, easier to miss. Every partial sum $S_N$ is a fraction, a ratio of two whole numbers, yet their limit $\pi^2/6$ is irrational (Section 9 says why). The partial sums close in on a number that none of them can ever equal.

Every solution of the problem comes down to finding where the circle is hiding. Euler found it in the zeros of the sine.

An engineer’s first instinct is to add terms until the answer settles and then recognise the number. Here that instinct fails, and the comparison in Proposition 2 tells us how badly. What is missing after $N$ terms, the tail $\sum_{n > N} 1/n^2$, is squeezed between two telescoping sums. Since

and the outer sums, taken over $n$ from $N + 1$ on, telescope to $1/(N+1)$ and $1/N$,

There is a picture behind this. The terms $1/n^2$ are the heights of the curve $y = 1/x^2$ at the whole numbers, so adding them is like adding the areas of rectangles of width 1 that sit under the curve. The area under the curve from $N$ to infinity is $\int_N^\infty dx/x^2 = 1/N$, and the rectangles fill most of it, so the tail is just under $1/N$.

The error after $N$ terms is therefore about $1/N$, and each further correct digit costs ten times as many terms. A thousand terms give 1.643935, of which only the first three digits are right (Table 1). Six correct decimals would take about two million terms. Adding terms and guessing was hopeless in the seventeenth century, and it would still be slow today.

The last column multiplies the error by $N$. It settles at 1, which is (3) at work: a hundred times more terms buy two more digits.

Euler found a way round this before he found the answer. In 1731 he computed the sum to six decimals, 1.644934, by turning the series into others that converge much faster [5]. The idea behind such accelerations can be seen in the tail. Instead of only trapping it between $1/(N+1)$ and $1/N$, we can describe it more and more exactly, as $1/N$ plus corrections in higher and higher powers of $1/N$:

The first term is the area under the curve from the picture above. Then comes a correction of half a term, the kind the trapezoid rule makes when it replaces a curve by straight segments, and the later terms correct the correction. This is an instance of what is now called the Euler–Maclaurin formula, which relates sums to integrals in general. Continued forever, the expansion would diverge, but its first few terms are remarkably accurate. The numbers $\frac16$ and $-\frac{1}{30}$ in it are Bernoulli numbers; they come back in Section 9.

Adding the first three corrections to just ten terms of the series gives 1.64493440, correct to six decimals, where the raw sum of ten terms is not correct to even one. Ten terms and a little algebra do the work of two million terms of brute force.

Digits alone, though, do not name a number. Six decimals narrow the possibilities without deciding between them, and nothing in 1.644934 announces a $\pi$, let alone a $\pi^2$. In hindsight the clue is there: multiply 1.644934 by 6 and take the square root, and out comes 3.141592…, the first seven digits of $\pi$. But nobody would try that particular pair of operations without already suspecting the answer. What the digits gave Euler was a target that any formula would have to hit. The formula itself needed an idea, and his came from the sine.

In 1735 Euler found the value [3] with an argument that takes a fact about polynomials and applies it where it has no right to apply. It gave him the answer, though not yet a proof.

A polynomial is determined, up to a constant factor, by its roots. If a polynomial $p$ of degree $d$ has the nonzero roots $r_1, \dots, r_d$ and takes the value 1 at $x = 0$, then

Each factor vanishes at one root and equals 1 at 0, so the product has the right roots and the right value at 0, and a polynomial of degree $d$ with $d$ given roots has no freedom left beyond a constant factor.

Multiplying out this product connects the roots with the coefficients. To get the term in $x$, take the $x$ from one factor and the 1 from every other; so the coefficient of $x$ is $-(1/r_1 + \cdots + 1/r_d)$, minus the sum of the reciprocals of the roots. With two roots, 2 and 3:

Knowing the roots of a polynomial tells us sums of reciprocals of its roots, and the Basel problem asks for exactly such a sum. The question becomes: which function has roots whose reciprocals, squared, are the numbers $1/n^2$?

The sine vanishes at every multiple of $\pi$: at 0, at $\pm\pi$, at $\pm 2\pi$, and so on. Dividing by $x$ removes the root at 0 and leaves a function that equals 1 there, since $\sin x / x$ approaches 1 as $x$ approaches 0. The function $\sin x / x$ therefore vanishes exactly at $x = \pm\pi, \pm 2\pi, \pm 3\pi, \dots$, and takes the value 1 at the origin, just like the polynomial $p$ above.

Euler treated $\sin x / x$ as if it were a polynomial of infinite degree with these roots. Pairing each root $n\pi$ with its negative $-n\pi$ gives a factor in $x^2$ alone,

and multiplying all of them together, he wrote

Here is the circle. The roots of the sine are the multiples of $\pi$, so the reciprocals of the squared roots are $\frac{1}{\pi^2} \cdot \frac{1}{n^2}$: our series, divided by $\pi^2$.

There is a second way to write $\sin x / x$ as a polynomial of infinite degree: its Taylor series. The Taylor series of a function at 0 is the power series whose coefficients match all the function’s derivatives there. The derivatives of the sine cycle through $\sin$, $\cos$, $-\sin$ and $-\cos$, whose values at 0 are 0, 1, 0 and $-1$, and so

for every $x$. Dividing by $x$,

Unlike the product, this series stands on solid ground; nobody doubts it. The trick is to put the two side by side.

Multiply out the product (5) and collect the terms in $x^2$. With only the first two factors,

Each further factor $1 - x^2/(n^2\pi^2)$ adds $-1/(n^2\pi^2)$ to the coefficient of $x^2$, because the only way to get a term in $x^2$ is to take it from one factor and 1 from all the others. With all the factors, the coefficient of $x^2$ is

In (6) the same coefficient is $-\frac16$. If the two expressions are the same function, their coefficients must agree, and setting them equal,

gives (2) at once. The value $\pi^2/6 = 1.644\,934\,066\ldots$ agreed with every digit Euler had computed.

As a proof, the argument has a gap, and it is a real one: a function is not determined by its zeros. Polynomials are, because a polynomial of degree $d$ has exactly $d$ roots and nothing else to adjust. Functions defined by infinite series have more freedom. The function $e^x \sin x / x$ vanishes at exactly the same points as $\sin x / x$, because $e^x$ is never zero, yet its Taylor series begins

with $\frac13$ where the sine has $-\frac16$. Euler’s argument, applied to this function, would give nonsense. So something special about the sine must rule out such an extra factor.

The explanation came only with Weierstrass’s factorization theorem of 1876 [6]. A function given by a power series that converges for every $x$ can be written as a product over its zeros, up to one extra factor of the form $e^{g(x)}$, which has no zeros at all. For the sine one can show that this extra factor is 1, and then (5) holds exactly as Euler wrote it.

Euler knew the gap was there. The digits convinced him, so he did what mathematicians still do with a result they believe but cannot yet prove: he kept looking for another way in.

The argument Euler wrote down in 1741 and published in 1743 [7] avoids infinite products altogether. It rests on one of the most useful habits in mathematics: computing the same quantity in two different ways, and reading off what the agreement says.

The quantity is an integral,

Computed one way, it is $\pi^2/8$. Computed the other way, term by term from a power series, it is the sum of the reciprocals of the odd squares, $1 + \frac19 + \frac{1}{25} + \cdots$. The two answers must be the same number, and from the odd squares the full sum follows in one line. To carry this out we need three facts: the derivative of the arcsine, its power series, and a family of integrals known today as Wallis integrals.

The arcsine undoes the sine: for $x$ between $-1$ and 1, $\arcsin x$ is the angle between $-\pi/2$ and $\pi/2$ whose sine is $x$. Differentiating the identity $\sin(\arcsin x) = x$ gives $\cos(\arcsin x) \cdot (\arcsin x)' = 1$, and the cosine of that angle is $\sqrt{1 - x^2}$, so

This settles the first computation. The integrand $\arcsin x / \sqrt{1 - x^2}$ is the derivative of $\frac12 (\arcsin x)^2$, so its integral from 0 to 1 is $\frac12 (\arcsin 1)^2 = \frac12 (\pi/2)^2 = \pi^2/8$.

For the second computation we need $\arcsin x$ as a power series. Its coefficients are built from the products of the odd and the even numbers, so it helps to give them a name:

The binomial series expands, for $|t| < 1$,

Put $t = x^2$ and integrate term by term from 0 to $x$. Since the left-hand side becomes the derivative of the arcsine, this gives, for $0 \le x < 1$,

Keep an eye on the products of odd and even numbers inside $c_n$: another pair of the same kind is about to appear, upside down.

The first few values are $I_0 = 1$, $I_1 = \frac23$, $I_2 = \frac{8}{15}$ and $I_3 = \frac{16}{35}$.

The cancellation in the middle is the heart of the proof, and small cases show it at work. For $n = 1$ the arcsine coefficient is $c_1 = \frac12$, divided by 3, and the Wallis integral is $I_1 = \frac23$, so the term is $\frac12 \cdot \frac13 \cdot \frac23 = \frac19$. For $n = 2$ it is $\frac38 \cdot \frac15 \cdot \frac{8}{15} = \frac{1}{25}$. The coefficients of the arcsine and the Wallis integrals are built from the same odd and even products, one upside down relative to the other, and they cancel almost completely, leaving only $1/(2n+1)^2$. Both come from the sine, the coefficients through its inverse and the integrals through its powers, so it is no surprise that they fit.

Why only the odd squares? Because the arcsine series contains only odd powers. That costs nothing. The even squares $\frac14 + \frac{1}{16} + \frac{1}{36} + \cdots$ are $\frac14$ of the whole sum, since $1/(2m)^2 = \frac14 \cdot 1/m^2$, so the odd squares are the other three quarters, and the last line of the proof undoes the split.

Every step used tools Euler’s readers already trusted: a power series, an integral and a substitution. There is no infinite product and no assumption about zeros. One step would be questioned today: integrating an infinite series term by term, which can go wrong in general. It was accepted practice then; the modern justification, that the terms are nonnegative (the monotone convergence theorem), came much later.

The proof below is Cauchy’s, from his Cours d’analyse of 1821 [8]; Proofs from THE BOOK devotes a chapter to this sum and its proofs [9]. It is a model of economy: no integrals and no infinite products, only trigonometry, the binomial theorem and one limit at the end. Its plan fits in two sentences. For a small angle $x$, the three numbers $\sin x$, $x$ and $\tan x$ are nearly equal, so $1/x^2$ is trapped between two trigonometric expressions. If we choose the angles cleverly, those expressions can be summed exactly, and then the trap closes on the partial sums of (1).

Draw a circle of radius 1 and an angle $x$ between 0 and $\pi/2$ at its centre. Three regions sit one inside the next: a triangle inside the sector, with area $\frac12 \sin x$; the sector of the circle, with area $\frac12 x$; and the right triangle whose far side touches the circle, with area $\frac12 \tan x$. Their areas are therefore in order,

Taking reciprocals reverses the inequalities, and squaring keeps them, so $1/\tan^2 x < 1/x^2 < 1/\sin^2 x$. In terms of the cotangent, $\cot x = \cos x / \sin x$, this reads

since $1/\sin^2 x = (\sin^2 x + \cos^2 x)/\sin^2 x = 1 + \cot^2 x$. Added up over well-chosen angles, the middle term will produce the Basel series, and what we need is the sum of $\cot^2 x$ over the same angles.

Why should such a sum be computable? The angles $k\pi/(2m+1)$, for $k = 1, \dots, m$, are exactly the points between 0 and $\pi/2$ where $\sin((2m+1)x)$ vanishes. De Moivre’s formula, $(\cos x + i \sin x)^n = \cos nx + i \sin nx$, writes $\sin nx$ as the imaginary part of $(\cos x + i \sin x)^n$, and expanding that power with the binomial theorem turns $\sin((2m+1)x)$, divided by a power of $\sin x$, into a polynomial in $\cot^2 x$. So the numbers $\cot^2(k\pi/(2m+1))$ are the roots of a polynomial we can write down.

Vieta’s formulas then give their sum without our computing a single cotangent. For a polynomial $a_m t^m + a_{m-1} t^{m-1} + \cdots$ with roots $t_1, \dots, t_m$, the roots add up to $-a_{m-1}/a_m$; this is the same bookkeeping as in Section 3, where the coefficient of $x$ recorded the sum of the reciprocals of the roots.

The lemma is easy to test. For $m = 1$ it says $\cot^2(\pi/3) = \frac13$, and indeed $\cot(\pi/3) = 1/\sqrt3$. For $m = 2$ it says $\cot^2(\pi/5) + \cot^2(2\pi/5) = 2$; numerically the two terms are 1.8944 and 0.1056. For $m = 3$ the three squared cotangents add up to exactly 5.

Why do both bounds close on $\pi^2/6$? For large $m$, the numerator $m(2m-1)$ behaves like $2m^2$ and the denominator $3(2m+1)^2$ like $12m^2$, so the lower bound approaches $\pi^2 \cdot 2/12 = \pi^2/6$, and the upper bound does the same. The upper bound is in fact exactly $\frac{\pi^2}{6}\bigl(1 - \frac{1}{(2m+1)^2}\bigr)$, so it approaches the limit quickly: for $m = 1000$ it falls short of $\pi^2/6$ by only $4 \times 10^{-7}$.

The proof needs no calculus beyond one limit at the end, no infinite products, and nothing from complex analysis except de Moivre’s formula. What it costs is some algebra, and a determined student can check every line of it.

In the nineteenth century Fourier showed how to write a function as a sum of sines and cosines [10], and his idea gives a proof that explains as much as it computes.

A Fourier series writes a function on the interval $(-\pi, \pi)$ as a combination of waves whose frequencies are whole numbers:

The coefficients measure how much of each wave the function contains, and they are computed by integrals:

An engineer would call them the spectrum of $f$: they say how strongly each frequency is present.

Why do those integrals pick out the coefficients? Because the waves are perpendicular to one another, in a sense that can be made precise. For two functions on $(-\pi, \pi)$, take $\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) g(x) \, dx$ as their inner product; it plays the role of the dot product of two vectors. For whole numbers $m, n \ge 1$, the identity $\sin mx \sin nx = \frac12\bigl(\cos(m-n)x - \cos(m+n)x\bigr)$ shows that

because the cosine of a nonzero whole multiple of $x$ integrates to zero over a full period. The same holds for the cosines, and every sine is perpendicular to every cosine. So the waves behave like unit vectors along the axes of a space with infinitely many directions, and the coefficient $b_n$ is the component of $f$ along $\sin nx$, found by an inner product exactly as the component of a vector is found by a dot product.

In the plane, a vector with coordinates $(3, 4)$ has length 5, because $3^2 + 4^2 = 5^2$. In space, the vector $(1, 2, 2)$ has length 3, because $1 + 4 + 4 = 9$. Whenever the axes are perpendicular, the square of the length is the sum of the squares of the coordinates. Parseval’s identity says the same thing for functions [11]: the squared length of $f$, measured by the integral of $f^2$, is the sum of the squares of its coordinates along the waves,

The $a_0^2/2$ comes from the constant term, whose own length is not 1; that is bookkeeping. The substance of the identity, and the real analysis in this proof, is that no length is lost in the infinite sum: the waves are enough to build every function whose square has a finite integral, so its coordinates account for all of it.

So the question becomes: which simple function has coordinates whose squares are $1/n^2$? The answer is the simplest function there is.

Repeated periodically, the function $x$ on $(-\pi, \pi)$ becomes a sawtooth: it climbs steadily, drops back, and climbs again. The proposition says that a sawtooth is built from sine waves whose strength falls off like $1/n$, and the squares of those strengths are, up to a factor 4, the terms of the Basel series.

Seen this way, the Basel sum is the squared length of a straight line, $f(x) = x$, measured in the coordinates of waves, and $\pi$ enters because waves are periodic with period $2\pi$. The whole weight of the proof rests on Parseval’s identity, which is where the real analysis lies. Applied to other functions, the same computation gives every even value $\zeta(2k)$; the polynomials that play the role of $x$ are the Bernoulli polynomials.

The last proof turns the sum into an area. It was found in 1983 in a version by Apostol [12], and in the form below by Beukers, Calabi and Kolk in 1993 [13]. The idea has two steps. First, write the sum as an integral over a square. Then find a change of variables that makes the integrand disappear, so that only an area is left.

The geometric series $1 + q + q^2 + \cdots = 1/(1 - q)$, valid for $|q| < 1$, turns a fraction into a sum. With $q = x^2 y^2$, where $x$ and $y$ lie between 0 and 1, it gives

and each term is easy to integrate over the unit square, because it splits into a function of $x$ times a function of $y$: the integral of $x^{2k} y^{2k}$ is $\int_0^1 x^{2k} dx \cdot \int_0^1 y^{2k} dy = \frac{1}{2k+1} \cdot \frac{1}{2k+1}$.

So the odd squares are an integral over a square. It remains to compute that integral.

An integral over a region can be computed in any coordinates, provided we account for how the new coordinates stretch area. The familiar example is polar coordinates: a small patch of size $dr \times d\theta$ at distance $r$ from the origin has area $r \, dr \, d\theta$, not $dr \, d\theta$, because arcs far from the centre are longer. In general, if $x$ and $y$ are functions of new variables $u$ and $v$, a small rectangle $du \times dv$ is carried to a small parallelogram whose area is $\bigl|\partial(x, y)/\partial(u, v)\bigr| \, du \, dv$, where

is the Jacobian determinant of the change of variables. For polar coordinates it is $r$.

Calabi’s substitution is designed so that this factor is exactly $1 - x^2 y^2$, the denominator of the integrand. The two cancel, and the integral becomes the area of the region we started from.

This time $\pi$ enters as the length of the legs of a right triangle, $\pi/2$, and the Basel sum is four thirds of that triangle’s area. Two of the proofs meet at the same number, $\pi^2/8$ for the odd squares, from opposite directions: Euler’s by an integral in one variable, this one by an area in two. Integrals of this kind reach further, too, as Section 9 shows.

Each route answers the question “where does $\pi$ come from?” differently. For Euler’s product, it is the spacing of the zeros of the sine. For the arcsine proof, it is the angle $\pi/2$ whose sine is 1. For Cauchy, it is the angles at which $\sin((2m+1)x)$ vanishes. For Parseval, it is the period of the waves; for the double integral, the legs of a triangle. It is the same circle every time, met from five directions.

In 1735 Euler had the answer and good reason to trust it. By 1741 he had a proof his contemporaries could check step by step. By today’s standards it was secured in the nineteenth century, when limit and convergence, which every one of these arguments quietly relies on, were finally made precise. And it has been proved again every few decades since, each new proof a small lesson in a different part of mathematics.

Euler’s method [3] also reaches every even value, because each further coefficient of (5) gives the next even sum, as the remark in Section 3 showed for $\zeta(4)$. In the form he later gave it,

where $B_2 = 1/6$, $B_4 = -1/30$, $B_6 = 1/42$, … are the Bernoulli numbers. They are the coefficients of the power series

and they turn up wherever sums of powers meet calculus: two of them already appeared in the tail expansion (4). So every even value is a rational number times a power of $\pi$. And since $\pi$ is transcendental, not a root of any polynomial with whole-number coefficients, as Lindemann proved in 1882 [14], every one of these values is irrational.

For the odd values, nothing of the kind is known, and not for want of trying. To prove a number irrational is to prove that no fraction $p/q$ equals it, however large $p$ and $q$ are, and without a formula to work from that is hard. The first real progress came only in 1978, when Roger Apéry announced that $\zeta(3) = 1.202\,056\,9\ldots$ is irrational; the proof appeared the next year [15]. His talk was so sketchy that much of the audience dismissed it, but Henri Cohen, Hendrik Lenstra and Alfred van der Poorten set out to check the argument, and within two months it stood. Van der Poorten’s account [16] is still the most engaging way into it, and Beukers soon gave a shorter proof with integrals close to (9) [17]. Even so, no formula for $\zeta(3)$ has been found, and nobody knows whether $\zeta(5)$ is irrational. Rivoal showed that infinitely many of the odd values are irrational [18], and Zudilin that at least one of $\zeta(5)$, $\zeta(7)$, $\zeta(9)$, $\zeta(11)$ is [19].

What has not been solved, then, is the same question one power up. Nobody has found a closed form for $1 + \frac{1}{8} + \frac{1}{27} + \cdots$, the sum of the reciprocals of the cubes, and after almost three centuries that silence is itself a kind of information: whatever made the even powers yield to Euler does not reach the odd ones.

The numbers above take a few lines to reproduce. Adding the terms from the smallest up keeps the rounding error of floating-point arithmetic well below the digits that matter: a computer stores about sixteen significant digits, and adding a tiny term to a large running total throws most of the tiny term’s digits away, so it pays to add the tiny terms together first.

/** The partial sum S_n, added from the smallest term up, which keeps rounding error small. */
function partialSum(n: number): number {
  let sum = 0;
  for (let k = n; k >= 1; k--) sum += 1 / (k * k);
  return sum;
}

/** Euler's estimate: S_n plus the first three terms of the tail. */
function eulerEstimate(n: number): number {
  return partialSum(n) + 1 / n - 1 / (2 * n ** 2) + 1 / (6 * n ** 3);
}

console.log(partialSum(1000)); // 1.6439345666815597
console.log(eulerEstimate(10)); // 1.6449343978332076
console.log(Math.PI ** 2 / 6); // 1.6449340668482264

Ten terms and three corrections beat a thousand terms by three digits, and a hundred terms with the same corrections agree with $\pi^2/6$ to eleven decimal places.

1. P. Mengoli, Novae quadraturae arithmeticae, Bologna, 1650.
2. J. Bernoulli, Tractatus de seriebus infinitis, appended to his Ars conjectandi, Basel, 1713. It collects the dissertations on series he published from 1686 on.
3. L. Euler, “De summis serierum reciprocarum,” Commentarii academiae scientiarum Petropolitanae 7 (1740), 123–134. Eneström index E41; facsimile in the Euler Archive.
4. R. Ayoub, “Euler and the zeta function,” The American Mathematical Monthly 81 (1974), 1067–1086. doi:10.1080/00029890.1974.11993738
5. L. Euler, “De summatione innumerabilium progressionum,” Commentarii academiae scientiarum Petropolitanae 5 (1738), 91–105. Written in 1731; Eneström index E20; facsimile in the Euler Archive.
6. K. Weierstrass, “Zur Theorie der eindeutigen analytischen Functionen,” Abhandlungen der Königlich Preussischen Akademie der Wissenschaften zu Berlin (1876).
7. L. Euler, “Démonstration de la somme de cette suite 1 + 1/4 + 1/9 + 1/16 + …,” Journal littéraire d’Allemagne, de Suisse et du Nord 2 (1743), 115–127. Written in 1741; Eneström index E63; facsimile in the Euler Archive.
8. A.-L. Cauchy, Cours d’analyse de l’École royale polytechnique, Paris, 1821, Note VIII.
9. M. Aigner and G. M. Ziegler, Proofs from THE BOOK, 6th ed., Springer, 2018, chapter “Three times π²/6.” doi:10.1007/978-3-662-57265-8
10. J. Fourier, Théorie analytique de la chaleur, Paris, 1822.
11. E. M. Stein and R. Shakarchi, Fourier Analysis: An Introduction, Princeton University Press, 2003.
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