Evaluating ζ(2)

Robin Chapman Department of Mathematics University of Exeter, Exeter, EX4 4QE, UK

                    rjc@maths.exeter.ac.uk

April 30, 1999

I list several proofs of the celebrated identity:

ζ(2) = ∑_{n=1}^{∞} \frac{1}{n^{2}} = \frac{π^{2}}{6} .  (1)

As it is clear that

\frac{3}{4}ζ(2) = ∑_{n=1}^{∞} \frac{1}{n^{2}} − ∑_{m=1}^{∞} \frac{1}{(2m)^{2}} = ∑_{r=0}^{∞} \frac{1}{(2r + 1)^{2}} ,

(1) is equivalent to

∑_{r=0}^{∞} \frac{1}{(2r + 1)^{2}} = \frac{π^{2}}{8} .  (2)

Many of the proofs establish this latter identity first.

None of these proofs is original; most are well known, but some are not as familiar as they might be. I shall try to assign credit the best I can, and I would be grateful to anyone who could shed light on the origin for any of these methods. I would like to thank Tony Lezard who spotted some errors in an earlier version, and Richard Carr for pointing out an egregious solecism.

Proof 1: Note that

\frac{1}{n^{2}} = ∫_{0}^{1}∫_{0}^{1} x^{n−1}y^{n−1}dx dy

and by the monotone convergence theorem we get

∑_{n=1}^{∞} \frac{1}{n^{2}} = ∫_{0}^{1}∫_{0}^{1} (∑_{n=1}^{∞}(xy)^{n−1}) dx dy
= ∫_{0}^{1}∫_{0}^{1} \frac{dx dy}{1 − xy} .

We change variables in this by putting (u, v) = ((x + y)/2, (y − x)/2), so that (x, y) = (u − v, u + v). Hence

ζ(2) = 2 ∫ ∫_{S} \frac{du dv}{1 − u^{2} + v^{2}}

where S is the square with vertices (0, 0), (1/2, −1/2), (1, 0) and (1/2, 1/2). Exploiting the symmetry of the square we get

ζ(2) = 4 ∫_{0}^{1/2}∫_{0}^{u} \frac{dv du}{1 − u^{2} + v^{2}} + 4 ∫_{1/2}^{1} ∫_{0}^{1−u} \frac{dv du}{1 − u^{2} + v^{2}}
= 4 ∫_{0}^{1/2} \frac{1}{\sqrt{1 − u^{2}}} tan^{−1} (\frac{u}{\sqrt{1 − u^{2}}} ) du

+4 ∫_{1/2}^{1} \frac{1}{\sqrt{1 − u^{2}}} tan^{−1} (\frac{1 − u}{\sqrt{1 − u^{2}}} ) du.

Now tan^{−1}(u/(\sqrt{1 − u^{2}})) = sin^{−1} u, and if θ = tan^{−1}((1 − u)/(\sqrt{1 − u^{2}})) then tan^{2} θ = (1 − u)/(1 + u) and sec^{2} θ = 2/(1 + u). It follows that u = 2 cos^{2} θ − 1 = cos 2θ and so θ = \frac{1}{2} cos^{−1} u = \frac{π}{4} − \frac{1}{2} sin^{−1} u. Hence

ζ(2) = 4 ∫_{0}^{1/2} \frac{sin^{−1} u}{\sqrt{1 − u^{2}}} du + 4 ∫_{1/2}^{1} \frac{1}{\sqrt{1 − u^{2}}} (\frac{π}{4} − \frac{sin^{−1} u}{2}) du
= [2(sin^{−1} u)^{2}]_{0}^{1/2} + [π sin^{−1} u − (sin^{−1} u)^{2}]_{1/2}^{1}
= \frac{π^{2}}{18} + \frac{π^{2}}{2} − \frac{π^{2}}{4} − \frac{π^{2}}{6} + \frac{π^{2}}{36}
= \frac{π^{2}}{6}

as required.

This is taken from an article in the Mathematical Intelligencer by Apostol in 1983.

Proof 2: We start in a similar fashion to Proof 1, but we use (2). We get

∑_{r=0}^{∞} \frac{1}{(2r + 1)^{2}} = ∫_{0}^{1}∫_{0}^{1} \frac{dx dy}{1 − x^{2}y^{2}} .

We make the substitution

(u, v) = (tan^{−1} x\sqrt{\frac{1 − y^{2}}{1 − x^{2}}} , tan^{−1} y\sqrt{\frac{1 − x^{2}}{1 − y^{2}}} )

so that

(x, y) = (\frac{sin u}{cos v} , \frac{sin v}{cos u}) .

The Jacobian matrix is

\frac{∂(x, y)}{∂(u, v)} = | \begin{matrix}cos u/ cos v & sin u sin v/ cos^{2} v \\ sin u sin v/ cos^{2} u & cos v/ cos u \end{matrix} |
= 1 − \frac{sin^{2} u sin^{2} v}{cos^{2} u cos^{2} v}
= 1 − x^{2}y^{2}.

Hence

\frac{3}{4}ζ(2) = ∫ ∫_{A} du dv

where

A = {(u, v) : u > 0, v > 0, u + v < π/2}

has area π^{2}/8, and again we get ζ(2) = π^{2}/6.

This is due to Calabi, Beukers and Kock.

Proof 3: We use the power series for the inverse sine function:

sin^{−1} x = ∑_{n=0}^{∞} \frac{1 · 3 ⋯ (2n − 1)}{2 · 4 ⋯ (2n)} \frac{x^{2n+1}}{2n + 1}

valid for |x| ≤ 1. Putting x = sin t we get

t = ∑_{n=0}^{∞} \frac{1 · 3 ⋯ (2n − 1)}{2 · 4 ⋯ 2n} \frac{sin^{2n+1} t}{2n + 1}

for |t| ≤ \frac{π}{2}. Integrating from 0 to \frac{π}{2} and using the formula

∫_{0}^{π/2} sin^{2n+1} x dx = \frac{2 · 4 ⋯ (2n)}{3 · 5 ⋯ (2n + 1)}

gives us

\frac{π^{2}}{8} = ∫_{0}^{π/2} t dt = ∑_{n=0}^{∞} \frac{1}{(2n + 1)^{2}}

which is (2).

This comes from a note by Boo Rim Choe in the American Mathematical Monthly in 1987.

Proof 4: We use the L^{2}-completeness of the trigonometric functions. Let e_{n}(x) = exp(2πinx) where n ∈ ℤ. The e_{n} form a complete orthonormal set in L^{2}[ 0, 1 ]. If we denote the inner product in L^{2}[ 0, 1 ] by ⟨ , ⟩, then Parseval’s formula states that

⟨f, f⟩ = ∑_{n=−∞}^{∞} |⟨f, e_{n}⟩|^{2}

for all f ∈ L^{2}[ 0, 1 ]. We apply this to f(x) = x. We easily compute ⟨f, f⟩ = \frac{1}{3}, ⟨f, e_{0}⟩ = \frac{1}{2} and ⟨f, e_{n}⟩ = \frac{1}{2πin} for n ≠ 0. Hence Parseval gives us

\frac{1}{3} = \frac{1}{4} + ∑_{n∈ℤ,n≠0} \frac{1}{4π^{2}n^{2}}

and so ζ(2) = π^{2}/6.

Alternatively we can apply Parseval to g = χ_{[0,1/2]}. We get ⟨g, g⟩ = \frac{1}{2}, ⟨g, e_{0}⟩ = \frac{1}{2} and ⟨g, e_{n}⟩ = ((−1)^{n} − 1)/2πin for n ≠ 0. Hence Parseval gives us

\frac{1}{2} = \frac{1}{4} + ∑_{r=0}^{∞} \frac{1}{π^{2}(2r + 1)^{2}}

and using (2) we again get ζ(2) = π^{2}/6.

This is a textbook proof, found in many books on Fourier analysis.

Proof 5: We use the fact that if f is continuous, of bounded variation on [ 0, 1 ] and f(0) = f(1), then the Fourier series of f converges to f pointwise. Applying this to f(x) = x(1 − x) gives

x(1 − x) = \frac{1}{6} + ∑_{n=1}^{∞} \frac{cos 2πnx}{π^{2}n^{2}},

and putting x = 0 we get ζ(2) = π^{2}/6. Alternatively putting x = 1/2 gives

\frac{π^{2}}{12} = − ∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}}

which again is equivalent to ζ(2) = π^{2}/6.

Another textbook proof.

Proof 6: Consider the series

f(t) = ∑_{n=1}^{∞} \frac{cos nt}{n^{2}}.

This is uniformly convergent on the real line. Now if ϵ > 0, then for t ∈ [ ϵ, 2π − ϵ] we have

∑_{n=1}^{N} sin nt = ∑_{n=1}^{N} \frac{e^{int} − e^{−int}}{2i}
= \frac{e^{it} − e^{i(N+1)t}}{2i(1 − e^{it})} − \frac{e^{−it} − e^{−i(N+1)t}}{2i(1 − e^{−it})}

and so this sum is bounded above in absolute value by

\frac{2}{|1 − e^{int}|} = \frac{2}{sin nt/2}.

Hence these sums are uniformly bounded on [ ϵ, 2π − ϵ] and by Dirichlet’s test the sum

∑_{n=1}^{∞} \frac{sin nt}{n}

is uniformly convergent on [ ϵ, 2π − ϵ]. It follows that for t ∈ (0, 2π)

f′(t) = − ∑_{n=1}^{∞} \frac{sin nt}{n}
= −Im (∑_{n=1}^{∞} \frac{e^{int}}{n} )
= Im(log(1 − e^{it}))
= arg(1 − e^{it})
= \frac{π − t}{2}.

By the fundamental theorem of calculus we have

f(π) − f(0) = ∫_{0}^{π} \frac{π − t}{2}dt = −\frac{π^{2}}{4} .

But f(0) = ζ(2) and f(−1) = ∑_{n=1}^{∞}(−1)^{n}/n^{2} = −ζ(2)/2. Hence ζ(2) = π^{2}/6. Alternatively we can put

D(z) = ∑_{n=1}^{∞} \frac{z^{n}}{n^{2}} ,

the dilogarithm function. This is uniformly convergent on the closed unit disc, and satisfies D′(z) = −(log(1 − z))/z on the open unit disc. Note that f(t) = D(e^{2πit}). We may now use arguments from complex variable theory to justify the above formula for f′(t).

This is just the previous proof with the Fourier theory eliminated.

Proof 7: We use the infinite product

sin πx = πx ∏_{n=1}^{∞} (1 − \frac{x^{2}}{n^{2}} )

for the sine function. Comparing coefficients of x^{3} in the MacLaurin series of sides immediately gives ζ(2) = π^{2}/6. An essentially equivalent proof comes from considering the coefficient of x in the formula

π cot πx = \frac{1}{x} + ∑_{n=1}^{∞} \frac{2x}{x^{2} − n^{2}} .

The original proof of Euler!

Proof 8: We use the calculus of residues. Let f(z) = πz^{−2} cot πz. Then f has poles at precisely the integers; the pole at zero has residue −π^{2}/3, and that at a non-zero integer n has residue 1/n^{2}. Let N be a natural number and let C_{N} be the square contour with vertices (±1 ± i)(N + 1/2). By the calculus of residues

−\frac{π^{2}}{3} + 2 ∑_{n=1}^{N} \frac{1}{n^{2}} = \frac{1}{2πi} ∫_{C_{N}} f(z) dz = I_{N}

say. Now if πz = x + iy a straightforward calculation yields

| cot πz|^{2} = \frac{cos^{2} x + sinh^{2} y}{sin^{2} x + sinh^{2} y} .

It follows that if z lies on the vertical edges of C_{n} then

| cot πz|^{2} = \frac{sinh^{2} y}{1 + sinh^{2} y} < 1

and if z lies on the horizontal edges of C_{n}

| cot πz|^{2} ≤ \frac{1 + sinh^{2} π(N + 1/2)}{sinh^{2} π(N + 1/2)} = coth^{2} π(N + 1/2) ≤ coth^{2} π/2.

Hence | cot πz| ≤ K = coth \frac{π}{2} on C_{N}, and so |f(z)| ≤ πK/(N +1/2)^{2} on C_{N}. This estimate shows that

|I_{n}| ≤ \frac{1}{2π} \frac{πK}{(N + 1/2)^{2}} 8(N + 1/2)

and so I_{N} → 0 as N → ∞. Again we get ζ(2) = π^{2}/6.

Another textbook proof, found in many books on complex analysis.

Proof 9: We first note that if 0 < x < \frac{π}{2} then sin x < x < tan x and so cot^{2} x < x^{−2} < 1 + cot^{2} x. If n and N are natural numbers with 1 ≤ n ≤ N this implies that

cot^{2} \frac{nπ}{(2N + 1)} < \frac{(2N + 1)^{2}}{n^{2}π^{2}} < 1 + cot^{2} \frac{nπ}{(2N + 1)}

and so

\frac{π^{2}}{(2N + 1)^{2}} ∑_{n=1}^{N} cot^{2} \frac{nπ}{(2N + 1)} < ∑_{n=1}^{N} \frac{1}{n^{2}}

< \frac{Nπ^{2}}{(2N + 1)^{2}} + \frac{π^{2}}{(2N + 1)^{2}} ∑_{n=1}^{N} cot^{2} \frac{nπ}{(2N + 1)}.

If

A_{N} = ∑_{n=1}^{N} cot^{2} \frac{nπ}{(2N + 1)}

it suffices to show that lim_{N→∞} A_{N}/N^{2} = \frac{2}{3}.

If 1 ≤ n ≤ N and θ = nπ/(2N + 1), then sin(2N + 1)θ = 0 but sin θ ≠ 0. Now sin(2N + 1)θ is the imaginary part of (cos θ + i sin θ)^{2N+1}, and so

\frac{sin(2N + 1)θ}{sin^{2N+1} θ} = \frac{1}{sin^{2N+1} θ} ∑_{k=0}^{N}(−1)^{k}\binom{2N + 1}{2N − 2k} cos^{2(N−k)} θ sin^{2k+1} θ

= ∑_{k=0}^{N}(−1)^{k}\binom{2N + 1}{2N − 2k} cot^{2(N−k)} θ
= f(cot^{2} θ)

say, where f(x) = (2N + 1)x^{N} − \binom{2N+1}{3}x^{N−1} + · · ·. Hence the roots of f(x) = 0 are cot^{2}(nπ/(2N + 1)) where 1 ≤ n ≤ N and so A_{N} = N(2N − 1)/3. Thus A_{N}/N^{2} → \frac{2}{3}, as required.

This is an exercise in Apostol’s Mathematical Analysis (Addison-Wesley, 1974).

Proof 10: Given an odd integer n = 2m + 1 it is well known that sin nx = F_{n}(sin x) where F_{n} is a polynomial of degree n. Since the zeros of F_{n}(y) are the values sin(jπ/n) (−m ≤ j ≤ m) and lim_{y→0}(F_{n}(y)/y) = n then

F_{n}(y) = ny ∏_{j=1}^{m} (1 − \frac{y^{2}}{sin^{2}(jπ/n)})

and so

sin nx = n sin x ∏_{j=1}^{m} (1 − \frac{sin^{2} x}{sin^{2}(jπ/n)}) .

Comparing the coefficients of x^{3} in the MacLaurin expansion of both sides gives

−\frac{n^{3}}{6} = −\frac{n}{6} − n ∑_{j=1}^{m} \frac{1}{sin^{2}(jπ/n)}

and so

\frac{1}{6} − ∑_{j=1}^{m} \frac{1}{n^{2} sin^{2}(jπ/n)} = \frac{1}{n^{2}} .

Fix an integer M and let m > M. Then

\frac{1}{6} − ∑_{j=1}^{M} \frac{1}{n^{2} sin^{2}(jπ/n)} = \frac{1}{n^{2}} + ∑_{j=M+1}^{m} \frac{1}{n^{2} sin^{2}(jπ/n)}

and using the inequality sin x > \frac{2}{π}x for 0 < x < \frac{π}{2}, we get

0 < \frac{1}{6} − ∑_{j=1}^{M} \frac{1}{n^{2} sin^{2}(jπ/n)} < \frac{1}{n^{2}} + ∑_{j=M+1}^{m} \frac{1}{4j^{2}} .

Letting m tend to infinity now gives

0 ≤ \frac{1}{6} − ∑_{j=1}^{M} \frac{1}{π^{2}j^{2}} ≤ ∑_{j=M+1}^{∞} \frac{1}{4j^{2}} .

Hence

∑_{j=1}^{∞} \frac{1}{π^{2}j^{2}} = \frac{1}{6}.

This comes from a note by Kortram in Mathematics Magazine in 1996. Proof 11: Consider the integrals

I_{n} = ∫_{0}^{π/2} cos^{2n} x dx and J_{n} = ∫_{0}^{π/2} x^{2} cos^{2n} x dx.

By a well-known reduction formula

I_{n} = \frac{1 · 3 · 5 ⋯ (2n − 1)}{2 · 4 · 6 ⋯ 2n} \frac{π}{2} = \frac{(2n)!}{4^{n}n!^{2}} \frac{π}{2} .

If n > 0 then integration by parts gives

I_{n} = [x cos^{2n} x]_{0}^{π/2} + 2n ∫_{0}^{π/2} x sin x cos^{2n−1} x dx
= n [x^{2} sin x cos^{2n−1} x]_{0}^{π/2} − n ∫_{0}^{π/2} x^{2}(cos^{2n} x − (2n − 1) sin^{2} x cos^{2n−2} x) dx
= n(2n − 1)J_{n−1} − 2n^{2}J_{n}.

Hence

\frac{(2n)!}{4^{n}n!^{2}} \frac{π}{2} = n(2n − 1)J_{n−1} − 2n^{2}J_{n}

and so

\frac{π}{4n^{2}} = \frac{4^{n−1}(n − 1)!^{2}}{(2n − 2)!} J_{n−1} − \frac{4^{n}n!^{2}}{(2n)!} J_{n}.

Adding this up from n = 1 to N gives

\frac{π}{4} ∑_{n=1}^{N} \frac{1}{n^{2}} = J_{0} − \frac{4^{N}N!^{2}}{(2N)!} J_{N}.

Since J_{0} = π^{3}/24 it suffices to show that lim_{N→∞} 4^{N}N!^{2}J_{N}/(2N)! = 0. But the inequality x < \frac{π}{2} sin x for 0 < x < \frac{π}{2} gives

J_{N} < \frac{π^{2}}{4} ∫_{0}^{π_{2}} sin^{2} x cos^{2N} x dx = \frac{π^{2}}{4} (I_{N} − I_{N+1}) = \frac{π^{2}I_{N}}{8(N + 1)}

and so

0 < \frac{4^{N}N!}{(2N)!}J_{N} < \frac{π^{3}}{16(N + 1)}.

This completes the proof.

This proof is due to Matsuoka (American Mathematical Monthly, 1961). Proof 12: Consider the well-known identity for the Fejér kernel:

(\frac{sin nx/2}{sin x/2})^{2} = ∑_{k=−n}^{n}(n − |k|)e^{ikx} = n + 2 ∑_{k=1}^{n}(n − k) cos kx.

Hence

∫_{0}^{π} x (\frac{sin nx/2}{sin x/2})^{2} dx = \frac{nπ^{2}}{2} + 2 ∑_{k=1}^{n}(n − k) ∫_{0}^{π} x cos kx dx
= \frac{nπ^{2}}{2} − 2 ∑_{k=1}^{n}(n − k)\frac{1 − (−1)^{k}}{k^{2}}
= \frac{nπ^{2}}{2} − 4n ∑_{1≤k≤n,2∤k} \frac{1}{k^{2}} + 4 ∑_{1≤k≤n,2∤k} \frac{1}{k}

If we let n = 2N with N an integer then

∫_{0}^{π} \frac{x}{8N} (\frac{sin Nx}{sin x/2})^{2} dx = \frac{π^{2}}{8} − ∑_{r=0}^{N−1} \frac{1}{(2r + 1)^{2}} + O (\frac{log N}{N} ) .

But since sin \frac{x}{2} > \frac{x}{π} for 0 < x < π then

∫_{0}^{π} \frac{x}{8N} (\frac{sin Nx}{sin x/2})^{2} dx < \frac{π^{2}}{8N} ∫_{0}^{π} sin^{2} Nx \frac{dx}{x} = \frac{π^{2}}{8N} ∫_{0}^{Nπ} sin^{2} y \frac{dy}{y} = O (\frac{log N}{N} ) .

Taking limits as N → ∞ gives

\frac{π^{2}}{8} = ∑_{r=0}^{∞} \frac{1}{(2r + 1)^{2}} .

This proof is due to Stark (American Mathematical Monthly, 1969). Proof 13: We carefully square Gregory’s formula

\frac{π}{4} = ∑_{n=0}^{∞} \frac{(−1)^{n}}{2n + 1}.

We can rewrite this as lim_{N→∞} a_{N} = \frac{π}{2} where

a_{N} = ∑_{n=−N}^{N} \frac{(−1)^{n}}{2n + 1}.

Let

b_{N} = ∑_{n=−N}^{N} \frac{1}{(2n + 1)^{2}} .

By (2) it suffices to show that lim_{N→∞} b_{N} = π^{2}/4, so we shall show that lim_{N→∞} (a_{N}^{2} − b_{N}) = 0.

If n ≠ m then

\frac{1}{(2n + 1)(2m + 1)} = \frac{1}{2(m − n)} (\frac{1}{2n + 1} − \frac{1}{2m + 1})

and so

a_{N}^{2} − b_{N} = ∑_{n=−N}^{N} ∑_{m=−N}^{N}′ \frac{(−1)^{m+n}}{2(m − n)} (\frac{1}{2n + 1} − \frac{1}{2m + 1})

= ∑_{n=−N}^{N} ∑_{m=−N}^{N}′ \frac{(−1)^{m+n}}{(2n + 1)(m − n)}

= ∑_{n=−N}^{N} \frac{(−1)^{n}c_{n,N}}{2n + 1}

where the dash on the summations means that terms with zero denominators are omitted, and

c_{n,N} = ∑_{m=−N}^{N}′ \frac{(−1)^{m}}{(m − n)}.

It is easy to see that c_{−n,N} = −c_{n,N} and so c_{0,N} = 0. If n > 0 then

c_{n,N} = (−1)^{n+1} ∑_{j=N−n+1}^{N+n} \frac{(−1)^{j}}{j}

and so |c_{n,N}| ≤ 1/(N − n + 1) as the magnitude of this alternating sum is not more than that of its first term. Thus

|a_{M}^{2} − b_{N}| ≤ ∑_{n=1}^{n} (\frac{1}{(2n − 1)(N − n + 1)} + \frac{1}{(2n + 1)(N − n + 1)})
= ∑_{n=1}^{n} \frac{1}{2N + 1} (\frac{2}{2n − 1} + \frac{1}{N − n + 1})

+ ∑_{n=1}^{n} \frac{1}{2N + 3} (\frac{2}{2n + 1} + \frac{1}{N − n + 1})

≤ \frac{1}{2N + 1}(2 + 4 log(2N + 1) + 2 + 2 log N)

and so a_{N}^{2} − b_{N} → 0 as N → ∞ as required.

This is an exercise in Borwein & Borwein’s Pi and the AGM (Wiley, 1987).

Proof 14: This depends on the formula for the number of representations of a positive integer as a sum of four squares. Let r(n) be the number of quadruples (x, y, z, t) of integers such that n = x^{2} + y^{2} + z^{2} + t^{2}. Trivially r(0) = 1 and it is well known that

r(n) = 8 ∑_{m|n,4∤m} m.

Let R(N) = ∑_{n=0}^{N} r(n). It is easy to see that R(N) is asymptotic to the volume of the 4-dimensional ball of radius \sqrt{N}, i.e., R(N) ∼ \frac{π^{2}}{2} N^{2}. But

R(N) = 1 + 8 ∑_{n=1}^{N} ∑_{m|n,4∤n} m = 1 + 8 ∑_{m≤N,4∤m} m ⌊\frac{N}{m} ⌋ = 1 + 8(θ(N) − 4θ(N/4))

where

θ(x) = ∑_{m≤x} m ⌊\frac{x}{m}⌋ .

But

θ(x) = ∑_{mr≤x} m

= ∑_{r≤x} ∑_{m=1}^{⌊x/r⌋} m

= \frac{1}{2} ∑_{r≤x} (⌊\frac{x}{r}⌋^{2} + ⌊\frac{x}{r}⌋)

= \frac{1}{2} ∑_{r≤x} (\frac{x^{2}}{r^{2}} + O (\frac{x}{r}))
= \frac{x^{2}}{2} (ζ(2) + O(1/x)) + O(x log x)
= \frac{ζ(2)x^{2}}{2} + O(x log x)

as x → ∞. Hence

R(N) ∼ \frac{π^{2}}{2} N^{2} ∼ 4ζ(2) (N^{2} − \frac{N^{2}}{4} )

and so ζ(2) = π^{2}/6.

This is an exercise in Hua’s book on number theory.