2 1. EARLY TRIUMPHS

Thus lim

N→∞

IN = 0, so from (1.2), we obtain

(1.3)

∞

n=1

1

n2k

= −

1

2

Res H(z)

1

z2k

, 0 .

To evaluate the right hand side of (1.3) explicitly, recall that the Bernoulli

numbers Bn are defined by

(1.4)

x

ex − 1

=

∞

=0

B x

!

.

In particular, B0 = 1, B1 = −1/2, B2 = 1/6, B4 = −1/30, B6 = 1/42, B8 = −1/30,

B10 = 5/66, B12 = −691/2730. Now from (1.4), we have

H(z) =

2πi

e2πiz − 1

=

∞

=0

B (2πi) z

−1

!

,

so that the coeﬃcient of 1/z in the Laurent expansion of

H(z)/z2k

about 0 is given

by

Res H(z)

1

z2k

, 0 =

(−1)kB2k(2π)2k

(2k)!

.

Plugging this into (1.3) yields

∞

n=1

1

n2k

=

(−1)k+122k−1

(2k)!

B2kπ2k,

which is the desired formula. In particular, taking k = 1, we have

ζ(2) =

∞

n=1

1

n2

=

π2

6

.

Comments. 1. Evaluating the sum

∑∞

n=1

1

n2

was a celebrated problem in the

mathematics of the late seventeenth and early eighteenth centuries. Originally

posed by Pietro Mengoli in 1644, it was brought to public attention by Jacob

Bernoulli in his Tractatus de Seriebus Infinitis (1689) and became known as the

Basel Problem. After many unsuccessful attempts by leading mathematicians, it

was finally solved in 1735 by Leonhard Euler, who produced a rigorous proof of the

result in 1741. Euler went on to discover the general formula for ζ(2k), evaluating

the sums explicitly for k up to 13. Of course, Euler’s arguments did not make use

of complex analysis, as that subject did not yet exist.

2. Expressing ζ(3) =

∞

∑

n=1

1

n3

in a simple closed form (or proving that no such

expression exists) remains an open problem of considerable interest; ditto for higher

odd powers. It is known (Apéry) that ζ(3) is an irrational number; for a proof,

see [B].

3. An extensive array of applications of the calculus of residues are displayed

in the two volumes [MK1], [MK2].