I'm trying to prove the following equality involving the Bernoulli polynomials:
$B_{k}(x)=N^{k-1}\sum_{a=0}^{N-1}B_{k}\left(\frac{x+a}{N}\right)$ for all $N\in\mathbb{N}$.
Since this is to be proved for all natural numbers, I immediately think that I need to use induction. Clearly the base case of $N=1$ is trivial. However, I'm not sure how to handle the general case. There are two ways that I've seen the Bernoulli polynomials presented:
A generating function: $\frac{te^{tx}}{e^{t}-1}=\sum_{k=0}^{\infty}B_{k}(x)\frac{t^{k}}{k!}.$
Using the Bernoulli numbers: we define $B_{k}(x)=\sum_{n=0}^{k}\binom{k}{n}b_{k-n}x^{n}$, where $b_{k-n}$ are the Bernoulli numbers.
I know that there are many other ways to define the Bernoulli polynomials. At any rate, would either of the definitions above be easily adopted to prove the desired equality?
I tried a bit already with the second definition by substituting $B_{k}\left(\frac{x+a}{N}\right)=\sum_{n=0}^{k}\binom{k}{n}b_{k-n}\left(\frac{x+a}{N}\right)^{n}$, but I quickly started to trip over notation. Thanks in advance for any suggestions.
Write $$C_k(x)=N^{k-1}\sum_{a=0}^{N-1}B_k\left(\frac{x+a}N\right).$$ Then \begin{align} \sum_{k=0}^\infty C_k(x)\frac{t^k}{k!} &=\frac1N\sum_{a=0}^{N-1}\sum_{k=0}^\infty B_k\left(\frac{x+a}N\right) \frac{(Nt)^k}{k!} =\frac1N\sum_{a=0}^{N-1}\frac{Nte^{t(x+a)}}{e^{Nt}-1}\\ &=\frac{e^{tx}}{e^{Nt}-1}\sum_{a=0}^{N-1}e^{at}=\frac{e^{tx}}{e^t-1} =\sum_{k=0}^\infty B_k(x)\frac{t^k}{k!}. \end{align}