Verifying a relation involving Bernoulli polynomials I would appreciate help, please, as to how to verify this relation from Kato's "Fermat's Dream" p.96.
He say: By the definition of $B_n(x)$, the Bernoulli polynomial, we have
$$\sum_{n=0}^{\infty}\frac{B_n(x)}{n!}u^n = \frac{u e^{xu}}{e^u - 1}$$
The definition for Bernoulli polynomials is, for $n \in \mathbb{N}$ is
$$B_n(x) = \sum_{k=0}^{n} \binom{n}{k}B_k x^{n - k}$$
I am trying
$$\sum_{n=0}^{\infty}\frac{B_n(x)}{n!}u^n = \sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{n}{k}B_k x^{n - k} u^n$$
Also for Bernoulli numbers there is,
$$\sum_{k=0}^{\infty}\frac{B_k}{n!} x^k = \frac{x}{e^x -1}$$
I don't know if what I am trying is on the right track. Either way, I would appreciate help pulling it all together. Thanks very much.
 A: Remember that the method of generating functions works with formal power series.  Using the generating series for the Bernoulli numbers and the exponential series we have:
$$\frac{u e^{xu}}{e^u - 1} = \frac{u}{e^u - 1} \cdot e^{xu}
= \left(\sum_{k=0}^{\infty}\frac{B_k}{k!} u^k\right) 
\left(\sum_{k=0}^{\infty}\frac{(xu)^k}{k!}\right)
$$
Now the rule for multiplication of power series gives:
\begin{align}\left(\sum_{k=0}^{\infty}\frac{B_k}{k!} u^k\right) 
\left(\sum_{k=0}^{\infty}\frac{x^k}{k!}u^k\right)
&= \sum_{n=0}^{\infty}\left( \sum_{k=0}^{n} \frac{B_k}{k!} \frac{x^{n-k}}{(n-k)!} \right) u^n \\
&= \sum_{n=0}^{\infty}\left(\frac{1}{n!} \sum_{k=0}^{n}n! \frac{B_k}{k!} \frac{x^{n-k}}{(n-k)!} \right) u^n \\
&= \sum_{n=0}^{\infty}\frac{ \sum_{k=0}^{n} {n \choose k}B_k x^{n-k}}{n!} u^n\\
&=:\sum_{n=0}^{\infty}\frac{B_n(x)}{n!}u^n
\end{align}
And from this you can read-off the formula for the Bernoulli polynomials
$$B_n(x) = \sum_{k=0}^{n} {n \choose k}B_k x^{n - k}$$
A: I am going to basically steal gammatester's answer and say it backwards, which I think is a much more intuitive way to look at things.
You are trying to compute $G(u) = \sum_{n=0}^\infty \frac{B_n(x)}{n!}u^n$, which is exactly the exponential generating function of the sequence $\{B_0 (x), B_1 (x), B_2(x), \ldots\}$ (we can, and should, treat $x$ as a formal symbol rather than a variable—it could just as well be $\pi$ in fact).
The sequence is defined by $B_n (x) = \sum_{k=0}^n {n\choose k} B_k x^{n-k}$.  But $c_n =\sum_{k=0}^n {n\choose k} a_k b_{n-k}$ is a type of convolution product of sequences which behaves behaves remarkably well with exponential generating functions.  Specifically, if $F$ and $G$ are the e.g.f.'s of $\{a_n\}$ and $\{b_n\}$, respectively, then $FG$ is the exponential generating function of $\{c_n\}$.
This means that we can reduce our problem to two much simpler problems:


*

*Find the exponential generating function of $\{x^n\}$.

*Find the exponential generating function of $\{B_n\}$.


But #1 is straightforward: $\sum_{n=0}^\infty \frac{x^n}{n!} u^n = e^{xu}$.  And #2 has already been given to you!—$\frac{u}{e^u-1}$ (make sure, in both cases, to remember that $u$, not $x$, is the variable for our functions).
So, an experienced generatingfunctionologist can look at this problem and instantly read off that the answer is the product of those two functions.
