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The Basel problem asks for the exact sum of the reciprocal squares of the positive integers. The answer isThe integral test proves that the series converges, but does not identify this exact value. The Euler evaluation of even zeta values does: the equally spaced poles of cotangent connect reciprocal integer powers to Bernoulli numbers and $\pi$. This separates three different questions: whether a sum exists, how well a partial sum approximates it, and what its exact value is.

For example, the ten-term partial sum is about $1.549768$, while the exact value is about $1.644934$. The zeta series truncation error bounds the missing amount between $1/11$ and $1/10$; it is much larger than the first omitted term $1/121$. More terms improve the approximation, but a numerical coincidence with ${\pi }^2/6$ would not prove the identity.

In On the sums of series of reciprocals by Leonhard Euler, §2 identifies the reciprocal-square sum with one-sixth of the squared circumference of a unit-diameter circle. The translation's publication note dates the presentation to 5 December 1735 and publication to 1740.^[ref] The cotangent coefficient proof of the even zeta values gives a short modern outline, while zeta series truncation error addresses the distinct approximation question.

The Euler evaluation of even zeta values turns a family of series into exact expressions using Bernoulli numbers, factorials and powers of $\pi$. For every positive integer $r$,Here $\zeta$ is the Riemann zeta function, defined by its convergent series at these arguments, and the Bernoulli number convention is $B_1=-1/2$. The alternating sign combines with the sign of $B_{2r}$ to give a positive sum.^[ref]

Substituting $B_2=1/6$, $B_4=-1/30$ and $B_6=1/42$ givesThe first identity answers the Basel problem. It is surprising because the original series only mentions reciprocal squares of integers, while its exact sum involves the circle constant $\pi$. The cotangent partial-fraction expansion supplies a connection between the two: trigonometric functions encode equally spaced integer poles, whose expansion brings the reciprocal powers into view.

For an original numerical comparison, the first ten terms give ${\sum }_{n=1}^{10}{n}^{-2}\approx 1.549767731$, while ${\pi }^2/6\approx 1.644934067$. The omitted positive terms still contribute approximately $0.095166336$. For fourth powers, the ten-term partial sum is approximately $1.082036583$, only $0.000286650$ below ${\pi }^4/90\approx 1.082323234$. Both are convergent series, but the second approaches its limit much faster.

Exact evaluation is useful beyond obtaining more digits. It supplies an independent check on a numerical summation, exposes how Bernoulli numbers organize infinitely many identities, and makes the remaining series truncation error measurable without knowing any omitted terms individually. The even integer condition is essential to this formula: putting $r=3/2$ into it is not a way to evaluate $\zeta (3)$, since the required Bernoulli number index and factorial expression here are indexed by positive integers.

Historically, Leonhard Euler announced reciprocal-square and reciprocal-fourth-power evaluations in §2 of On the sums of series of reciprocals by Leonhard Euler. That paper was presented on 5 December 1735 and published in 1740, as the English translation's first-page note records. His §§3–8 develop a route through the sine power series and relationships between roots and coefficients. The cotangent calculation below is a concise modern presentation of the connection.^[ref]

For the full modular group $\Gamma =S{L}_2(\mathbb{Z})$ and integer weight $k$, a modular form is a holomorphic function on the upper half-plane satisfyingand holomorphic at the cusp at infinity. Explicitly, periodicity under $\tau \mapsto \tau +1$ gives a Fourier expansion of a modular form in $q=e^{2\pi i\tau }$, and holomorphy at the cusp means $f={\sum }_{n\ge 0}{a}_n{q}^n$. A cusp form has $a_0=0$. All cusps are equivalent for this group. The matrix $-I$ forces $f=(-1{)}^{k}f$, so every odd-weight modular form is zero.

There is a corresponding qualification to the displayed Eisenstein series formula: the Bernoulli-normalized formula is for even integers $k\ge 4$. For odd integers $k>2$, pairing $(m,n)$ with $(-m,-n)$ gives $G_k=0$, while $B_k=0$, so the printed expression divided by $B_k$ is undefined. We prove the asserted nonzero formula in its even-weight range.

On each compact set in the upper half-plane, the real-linear map $(m,n)\mapsto m\tau +n$ has a uniformly bounded inverse. Hence $|m\tau +n|\ge C(m^2+n^2{)}^{1/2}$ there. For $k>2$, lattice convergence gives locally uniform absolute convergence of $G_k$, and therefore holomorphy. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma$, thenThe map $(m,n)\mapsto (ma+nc,mb+nd)$ permutes the nonzero integer pairs. Reindexing proves the weight-$k$ transformation law.

To compute the Fourier expansion of a normalized Eisenstein series, start from the cotangent partial-fraction expansionFor $\mathrm{Im}z>0$, the cotangent Fourier expansion is $\pi \cot (\pi z)=-\pi i-2\pi i{\sum }_{r\ge 1}{e}^{2\pi irz}$. Differentiating both normally convergent expressions $k-1$ times givesFor even $k$, the terms with $m=0$ sum to $2\zeta (k)$, and the terms with $m<0$ duplicate those with $m>0$. Substituting $z=m\tau$ in the last identity and collecting $N=mr$ givesThe rearrangement is absolutely convergent for every $|q|<1$, since the divisor function grows at most polynomially.

For completeness, the normalization follows directly from the definition of the Bernoulli numbers. Near zero, the cotangent partial-fraction expansion givesOn the other hand $\pi \cot (\pi z)=\pi i+2\pi i/(e^{2\pi iz}-1)$. Substitute $t=2\pi iz$ in the generating function of the Bernoulli numbers. The term $B_1=-1/2$ cancels $\pi i$, and the higher odd terms vanish because $t/(e^t-1)+t/2$ is even. Comparison of Laurent coefficients proves the Euler evaluation of even zeta valuesThusThis Fourier series also proves holomorphy at infinity, completing the modular form verification.

For the Ramanujan congruence modulo 691, the standard facts we use are that the modular discriminant is the weight-twelve cusp form $\Delta =q{\prod }_{n\ge 1}(1-q^n{)}^{24}={\sum }_{n\ge 1}\tau (n)q^n$, and that $S_{12}=\mathbb{C}\Delta$. The latter also follows from the valence formula for the modular group: a nonzero weight-twelve modular form has total weighted zero order one, so a cusp form whose coefficient of $q$ vanishes would have cusp order at least two and must be zero. The product makes the Ramanujan tau function integer-valued.

Write $E_k=G_k/(2\zeta (k))$. Using $B_4=-1/30$ and the supplied $B_{12}=-691/2730$, we haveBoth $E_{12}$ and $E_4^3$ have weight twelve and constant term one. Their difference is a cusp form, so is a multiple of $\Delta$. Its coefficient of $q$ is $65520/691-720=-432000/691$, which determines that multiple:The coefficients of $E_4^3$ are integers. Comparing the positive-index Fourier coefficients and reducing modulo $691$ therefore gives $65520{\sigma }_{11}(n)\equiv -432000\tau (n)$. Both numerical constants are congruent to $566$ modulo $691$, and $\gcd (566,691)=1$. Cancellation proves