Singularities/Residues/Applications 737
corresponding Bernoulli number is known. For instance, since B 2 = 1/ 6
we have
rr^2 1 1 1
6 = 1 + 22 + 32 + 42 + ... (9.12-7)
Similarly, since B4 = -% 0 we find that
rr^4 1 1 1
90 =^1 + 2 4 + 3 4 + 4 4 + ... (9.12-8)
The reader will note that (9.12-6) implies that (-l)n-l B 2 n > 0: i.e., the
Bernoulli numbers B 2 n ( n ;:::: 1) are alternatively positive and negative.
4. Let f(z) = rrcscrrz and g(z) = 1/z^2 n. Replacing z by rrz in (8.5-8),
we have
00
(2^2 n - 2)B2
rrzcscrrz = 1 + L(-1t-^1 n rr^2 nz^2 n
n=l (2n)!
for lzl < 1. From this expansion it follows that
Resrrzcscrrz · z-(^2 n+l) = (-lt-^1 (2^2 n - 2)rr^2 n ~ B
z=O (2n)!
and since /3k = Resz=k rr csc rrz = (-l)k we have, by (9.12-2),
+co ( )k +oo ( )k+l
(-l)n-1(^2 2n _ 2)rr2n (2n B2n )! = _ '"'' L.J ~ k2n = '"'' L.J _-_1_ k2n
k=-oo k=-oo
which can be written as
~(22n2)rr2n (-l)n-lB2n = ~ (-l)k+l = 1-!__+_1 __ ··· (9.12-9)
2 (2n)! ~ k2n 22n. 32n
This equation gives another representation of the Bernoulli numbers, this
time in terms of an alternating p-series. Or we may use (9.12-9) to evalu-
ate the p-series (p even) if the corresponding Bernoulli number is already ,
known. For instance, for n = 1 we find that.
rr^2 1 1 1
12 =^1 - 2 2 + 3 2 - 4 2 + ... (9.12-10)
and for n = 2,
7rr^4 1 · 1 1
-=1--+---+···
720 24 34 44
(9.12-11)