I am trying to show that $B_n(x+1)-B_n(x)=nx^{n-1}$, where $B_n(x)$ is the Bernoulli polynomial. In order to avoid circular reasoning, I define the Bernoulli's number $B_n$ as the coefficient in the following generating function:
$$\frac{x}{e^x-1}=\sum_{n=0}^\infty \frac{B_n}{n!}x^n$$
and define the Bernoulli's polynomial as
$$B_n(x)=\sum_{k=0}^n \begin{pmatrix}n\\ k\end{pmatrix}B_{n-k}x^k$$
From the above definition, I am able to derive the translational property:
$$B_n(x+y)=\sum_{r=0}^n \begin{pmatrix}n\\ r\end{pmatrix}B_r(x)y^{n-r}$$
Then by putting $y=1$, we have
\begin{align*}
B_n(x+1)-B_n(x)&=\sum_{r=0}^n\begin{pmatrix}n\\ r\end{pmatrix}B_r(x)-B_n(x) \\
&= \sum_{r=0}^{n-1}\begin{pmatrix}n\\ r\end{pmatrix}B_r(x) \\
&= \sum_{r=0}^{n-1}\sum_{k=0}^r\begin{pmatrix}n\\ r\end{pmatrix}\begin{pmatrix}r\\ k\end{pmatrix}B_{r-k}x^k \\
&= \sum_{k=0}^{n-1}\sum_{r=k}^{n-1}\begin{pmatrix}n\\ r\end{pmatrix}\begin{pmatrix}r\\ k\end{pmatrix}B_{r-k}x^k \\
&= \sum_{k=0}^{n-1}x^k\sum_{r=k}^{n-1}\begin{pmatrix}n\\ r\end{pmatrix}\begin{pmatrix}r\\ k\end{pmatrix}B_{r-k} \\
\end{align*}
It is easy to see that $\displaystyle \sum_{r=k}^{n-1}\begin{pmatrix}n\\ r\end{pmatrix}\begin{pmatrix}r\\ k\end{pmatrix}B_{r-k}=n$ when $k=n-1$, but I have no idea how to show that this is equal to $0$ when $k$ is not $n-1$. Any help would be appreciated.
Proving Property of Bernoulli’s polynomial $B_n(x+1)-B_n(x)=nx^{n-1}$
bernoulli numbersbernoulli-polynomials
Best Answer
From your main definition of the Bernoulli numbers, for $ x\in\mathcal{B}\left(0,2\pi\right) $, multiplying both sides by $ \mathrm{e}^{x} $, we have : \begin{aligned}\frac{x\,\mathrm{e}^{x}}{\mathrm{e}^{x}-1}&=\left(\sum_{n=0}^{+\infty}{\frac{B_{n}}{n!}x^{n}}\right)\left(\sum_{n=0}^{+\infty}{\frac{x^{n}}{n!}}\right)\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{-x}{\mathrm{e}^{-x}-1}&=\sum_{n=0}^{+\infty}{\frac{1}{n!}\left(\sum_{k=0}^{n}{\binom{n}{k}B_{k}}\right)x^{n}}\\ \iff\sum_{n=0}^{+\infty}{\frac{\left(-1\right)^{n}B_{n}}{n!}x^{n}}&=\sum_{n=0}^{+\infty}{\frac{1}{n!}\left(\sum_{k=0}^{n}{\binom{n}{k}B_{k}}\right)x^{n}}\end{aligned}
Thus, for all $ n\in\mathbb{N} $, we have : $$ \sum_{k=0}^{n}{\binom{n}{k}B_{k}}=\left(-1\right)^{n}B_{n} $$
Now : \begin{aligned} \sum_{r=k}^{n-1}{\binom{n}{r}\binom{r}{k}B_{r-k}}&=\binom{n}{k}\sum_{r=k}^{n-1}{\binom{n-k}{r-k}B_{r-k}}\\ &=\binom{n}{k}\sum_{r=0}^{n-k-1}{\binom{n-k}{r}B_{r}}\\ &=\binom{n}{k}\sum_{r=0}^{n-k}{\binom{n-k}{r}B_{r}}-\binom{n}{k}B_{n-k}\\ &=\binom{n}{k}\left(-1\right)^{n-k}B_{n-k}-\binom{n}{k}B_{n-k}\\ \sum_{r=k}^{n-1}{\binom{n}{r}\binom{r}{k}B_{r-k}}&=\left(\left(-1\right)^{n-k}-1\right)\binom{n}{k}B_{n-k}\end{aligned}
Let $ k\in\left[\!\left[0,n-2\right]\!\right] $, If $ k $ is such that $ n-k $ is even than $ \left(-1\right)^{n-k}-1 =0 $, but if $ k $ is such that $ n-k $ is odd than $ B_{n-k}=0 $, because besides $ B_{1} $, all $ B_{n} $ such that $ n $ is odd are equal to $ 0 $. Thus, in all cases $ \left(\left(-1\right)^{n-k}-1\right)B_{n-k}=0 $ for all $ k\in\left[\!\left[0,n-2\right]\!\right] $. Hence the result.