I'm reading Tenenbaum's Introduction to analytic number theory.
He defines Bernoulli polynomials as the unique sequence $B_n$ such that
$B_0=1$
$\forall n\geq0, B_{n+1}'(X)=(n+1)B_n(X)$
$\forall n\geq 1, \int_0^1 B_n(t) dt=0$
Tenenbaum writes
One easily verifies that these assumptions imply the identity
$\displaystyle \sum_{r=0}^\infty B_r(x) \frac{y^r}{r!} = \frac{ye^{xy}}{e^y-1}$
(for fixed $x$)
How is that easily verified ?
For fixed $x$, I tried to write $B_{n+1}'(x)y^r=(r+1)B_r(x)y^r$ and then sum over $r$. $\sum_r (r+1)B_r(x)y^r$ is the derivative of some power series, but I can do nothing with $\sum_r B_{n+1}'(x)y^r$.
Denote by $T(x)$ the LHS, so that $$T(x)=\sum_{r=0}^{\infty} B_r(x)\frac{y^r}{r!}=1+\sum_{r=1}^{\infty} B_r(x)\frac{y^r}{r!} \tag{1}$$
Differentiating (1) with respect to $x$, we obtain $$T'(x)=\sum_{r=1}^{\infty} rB_{r-1}(x)\frac{y^r}{r!}= y\sum_{r=1}^{\infty} B_{r-1}(x)\frac{y^{r-1}}{(r-1)!}=yT(x)$$.
This is a classical linear homogeneous differnetial equation of order $1$, so there is a univariate function $f$ such that $$T(x)=f(y)e^{xy}\tag{2}$$.
Integrating (1) with respect to $x$ between $0$ and $1$, we see that $\int_{0}^{1}T(x)dx=1$. If we integrate (2) on the other hand, we obtain $\int_{0}^{1}T(x)dx=f(y)\frac{e^{y}-1}{y}$, so $f(y)=\frac{y}{e^y-1}$ which concludes the computation.