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].
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