# Basel problem

The Basel problem is a problem in mathematical analysis with relevance to number theory, first posed by Pietro Mengoli in 1650 and solved by Leonhard Euler in 1734[1] and read on 5 December 1735 in The Saint Petersburg Academy of Sciences.[2] Since the problem had withstood the attacks of the leading mathematicians of the day, Euler's solution brought him immediate fame when he was twenty-eight. Euler generalised the problem considerably, and his ideas were taken up years later by Bernhard Riemann in his seminal 1859 paper "On the Number of Primes Less Than a Given Magnitude", in which he defined his zeta function and proved its basic properties. The problem is named after Basel, hometown of Euler as well as of the Bernoulli family who unsuccessfully attacked the problem.

The Basel problem asks for the precise summation of the reciprocals of the squares of the natural numbers, i.e. the precise sum of the infinite series:

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\cdots }$ .

The sum of the series is approximately equal to 1.644934.[3] The Basel problem asks for the exact sum of this series (in closed form), as well as a proof that this sum is correct. Euler found the exact sum to be π2/6 and announced this discovery in 1735. His arguments were based on manipulations that were not justified at the time, although he was later proven correct, and it was not until 1741 that he was able to produce a truly rigorous proof.

## Euler's approach

Euler's original derivation of the value π2/6 essentially extended observations about finite polynomials and assumed that these same properties hold true for infinite series.

Of course, Euler's original reasoning requires justification (100 years later, Karl Weierstrass proved that Euler's representation of the sine function as an infinite product is valid, by the Weierstrass factorization theorem), but even without justification, by simply obtaining the correct value, he was able to verify it numerically against partial sums of the series. The agreement he observed gave him sufficient confidence to announce his result to the mathematical community.

To follow Euler's argument, recall the Taylor series expansion of the sine function

${\displaystyle \sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots }$

Dividing through by x, we have

${\displaystyle {\frac {\sin x}{x}}=1-{\frac {x^{2}}{3!}}+{\frac {x^{4}}{5!}}-{\frac {x^{6}}{7!}}+\cdots }$

Using the Weierstrass factorization theorem, it can also be shown that the left-hand side is the product of linear factors given by its roots, just as we do for finite polynomials (which Euler assumed as a heuristic for expanding an infinite degree polynomial in terms of its roots, but is in general not always true for general ${\displaystyle P(x)}$):[4]

{\displaystyle {\begin{aligned}{\frac {\sin x}{x}}&=\left(1-{\frac {x}{\pi }}\right)\left(1+{\frac {x}{\pi }}\right)\left(1-{\frac {x}{2\pi }}\right)\left(1+{\frac {x}{2\pi }}\right)\left(1-{\frac {x}{3\pi }}\right)\left(1+{\frac {x}{3\pi }}\right)\cdots \\&=\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots \end{aligned}}}

If we formally multiply out this product and collect all the x2 terms (we are allowed to do so because of Newton's identities), we see by induction that the x2 coefficient of sin x/x is [5]

${\displaystyle -\left({\frac {1}{\pi ^{2}}}+{\frac {1}{4\pi ^{2}}}+{\frac {1}{9\pi ^{2}}}+\cdots \right)=-{\frac {1}{\pi ^{2}}}\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}.}$

But from the original infinite series expansion of sin x/x, the coefficient of x2 is 1/3! = −1/6. These two coefficients must be equal; thus,

${\displaystyle -{\frac {1}{6}}=-{\frac {1}{\pi ^{2}}}\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}.}$

Multiplying both sides of this equation by −π2 gives the sum of the reciprocals of the positive square integers.

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}.}$

This method of calculating ${\displaystyle \zeta (2)}$ is detailed in expository fashion most notably in Havil's Gamma book which details many zeta function and logarithm-related series and integrals, as well as a historical perspective, related to the Euler gamma constant.[6]

### Generalizations of Euler's method using elementary symmetric polynomials

Using formulas obtained from elementary symmetric polynomials,[7] this same approach can be used to enumerate formulas for the even-indexed even zeta constants which have the following known formula expanded by the Bernoulli numbers:

${\displaystyle \zeta (2n)={\frac {(-1)^{n-1}(2\pi )^{2n}}{2\cdot (2n)!}}B_{2n}.}$

For example, let the partial product for ${\displaystyle \sin(x)}$ expanded as above be defined by ${\displaystyle {\frac {S_{n}(x)}{x}}:=\prod \limits _{k=1}^{n}\left(1-{\frac {x^{2}}{k^{2}\cdot \pi ^{2}}}\right)}$. Then using known formulas for elementary symmetric polynomials (a.k.a., Newton's formulas expanded in terms of power sum identities), we can see (for example) that

{\displaystyle {\begin{aligned}\left[x^{4}\right]{\frac {S_{n}(x)}{x}}&={\frac {1}{2\pi ^{4}}}\left(\left(H_{n}^{(2)}\right)^{2}-H_{n}^{(4)}\right)\qquad {\xrightarrow {n\rightarrow \infty }}\qquad {\frac {1}{2}}\left(\zeta (2)^{2}-\zeta (4)\right)\\&\qquad \implies \zeta (4)={\frac {\pi ^{4}}{90}}=-2\pi ^{2}\cdot [x^{4}]{\frac {\sin(x)}{x}}+{\frac {\pi ^{4}}{36}}\\\left[x^{6}\right]{\frac {S_{n}(x)}{x}}&=-{\frac {1}{6\pi ^{6}}}\left(\left(H_{n}^{(2)}\right)^{3}-2H_{n}^{(2)}H_{n}^{(4)}+2H_{n}^{(6)}\right)\qquad {\xrightarrow {n\rightarrow \infty }}\qquad {\frac {1}{6}}\left(\zeta (2)^{3}-3\zeta (2)\zeta (4)+2\zeta (6)\right)\\&\qquad \implies \zeta (6)={\frac {\pi ^{6}}{945}}=-3\cdot \pi ^{6}[x^{6}]{\frac {\sin(x)}{x}}-{\frac {2}{3}}{\frac {\pi ^{2}}{6}}{\frac {\pi ^{4}}{90}}+{\frac {\pi ^{6}}{216}},\end{aligned}}}

and so on for subsequent coefficients of ${\displaystyle [x^{2k}]{\frac {S_{n}(x)}{x}}}$. There are other forms of Newton's identities expressing the (finite) power sums ${\displaystyle H_{n}^{(2k)}}$ in terms of the elementary symmetric polynomials, ${\displaystyle e_{i}\equiv e_{i}\left(-{\frac {\pi ^{2}}{1^{2}}},-{\frac {\pi ^{2}}{2^{2}}},-{\frac {\pi ^{2}}{3^{2}}},-{\frac {\pi ^{2}}{4^{2}}},\cdots \right),}$ but we can go a more direct route to expressing non-recursive formulas for ${\displaystyle \zeta (2k)}$ using the method of elementary symmetric polynomials. Namely, we have a recurrence relation between the elementary symmetric polynomials and the power sum polynomials given as on this page by

${\displaystyle (-1)^{k}ke_{k}(x_{1},\ldots ,x_{n})=\sum _{j=1}^{k}(-1)^{k-j-1}p_{j}(x_{1},\ldots ,x_{n})e_{k-j}(x_{1},\ldots ,x_{n}),}$

which in our situation equates to the limiting recurrence relation (or generating function convolution, or product) expanded as

${\displaystyle {\frac {\pi ^{2k}}{2}}\cdot {\frac {(2k)\cdot (-1)^{k}}{(2k+1)!}}=-[x^{2k}]{\frac {\sin(\pi x)}{\pi x}}\times \sum _{i\geq 1}\zeta (2i)x^{i}.}$

Then by differentiation and rearrangement of the terms in the previous equation, we obtain that

${\displaystyle \zeta (2k)=[x^{2k}]{\frac {1}{2}}\left(1-\pi x\cot(\pi x)\right).}$

### Consequences of Euler's proof

By Euler's proof for ${\displaystyle \zeta (2)}$ explained above and the extension of his method by elementary symmetric polynomials in the previous subsection, we can conclude that ${\displaystyle \zeta (2k)}$ is always a rational multiple of ${\displaystyle \pi ^{2k}}$. Thus compared to the relatively unknown, or at least unexplored up to this point, properties of the odd-indexed zeta constants, including Apéry's constant ${\displaystyle \zeta (3)}$, we can conclude much more about this class of zeta constants. In particular, since ${\displaystyle \pi }$ and integer powers of it are transcendental, we can conclude at this point that ${\displaystyle \zeta (2k)}$ is irrational, and more precisely, transcendental for all ${\displaystyle k\geq 1}$.

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