## Monday Math 57

Previously, we introduced the Bernoulli polynomials, and the Bernoulli numbers. In particular, we found that the Bernoulli numbers obey (and along with B0=1 can be defined by) the relation .

Now, let us examine the Maclaurin series for the function  (the singularity at z=0 is removable, and is also removable in the derivatives of all order).
Let , so that our series is then

Now, the Maclaurin series for ex is , so the series for ez-1 is

Thus, we have:
.
Now, we use the Cauchy product for infinite series:
, where .
This tells us
,
where

But the product of the two series is unity, so we have c0=1, cn=0 for n≥1. Thus

and
 for n≥1
This, however, is the relation for the Bernoulli numbers: , and so the Maclaurin series is
,
and .