We first consider the generating function for Bernoulli numbers (
):
and therefore:
_{z=0})
We can easily calculate the first three Bernoulli numbers and the remaining numbers can be calculated via contour integration and expressed in terms of the zeta function at the negative integers. We find:
where all even Bernoulli numbers are non-zero and
for
. Now, add
to both sides of the generating expression:
+B_0+B_1 z+B_2\frac{z^2}{2}+\cdots))
but
and the remaining odd Bernoulli numbers are zero. Thus:
)
We can then write the expression as:
!})
Now note the left size can be written as:
=\frac{z}{2}\left(\frac{e^{z/2}+e^{-z/2}}{e^{z/2}-e^{-z/2}}\right))
so therefore:
=\sum_{n=0}^{\infty}\ frac{B_{2n}z^{2n}}{(2n)!})
Letting
then:
=\sum_{n=0}^{\infty}\frac{B_{2n}(2iw)^{ 2n}}{(2n)!})
which simplifies to:
=\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} 2^{2n} w^{2n}}{(2n)!})
or:
=\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} 2^{2n}w^{2n-1}}{(2n)!})
we now use the identity:
-2\cot(2w)=\tan(w))
and therefore:
=\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} 2^{2n} w^{2n-1}}{(2n)!}-2\sum_{n=0}^{\infty}\frac{(-)^n B_{2n}(2n)^{2n-1}w^{2n-1}}{(2n)!})
which then simplifies to:
=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} 2^{2n}\left(2^{2n}-1\right) B_{2n} w^{2n-1}}{(2n)!},\quad -\pi/2<w<\pi/2)
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About LinkBacks=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} 2^{2n}(2^{2n}-1) B_{2n}}{(2n)!} z^{2n-1})
