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\title{Several explicit formulae for Bernoulli polynomials}

\markboth{T.\,Komatsu and C.\,De J.\,Pita Ruiz V.}{Several explicit formulae for Bernoulli polynomials}

\author[T.\,Komatsu and C.\,De J.\,Pita Ruiz V.]{Takao Komatsu\affil{1}\comma\corrauth and Claudio De J.\,Pita Ruiz V.\affil{2}}

\address{\affilnum{1}\ School of Mathematics and Statistics, Wuhan University, Wuhan 430\,072, China\\
\affilnum{2}\ Universidad Panamericana, Calle Augusto Rodin 498, Benito Ju\'{a}rez, Insurgentes Mixcoac, 03920 Ciudad de M\'{e}xico, D.\,F., Mexico}

\emails{{\tt komatsu@whu.edu.cn} (T.\,Komatsu), {\tt cpita@up.edu.mx} (C.\,De J.\,Pita Ruiz V.)}


\begin{abstract}
We prove several explicit formulae for the $n$-th Bernoulli polynomial $B_{n}(x)$, in which $B_{n}(x)$ is equal to an affine combination of the polynomials $(x-1)^{n}$, $(x-2)^{n}, \dots, (x-k-1)^{n}$, where $k$ is any fixed positive integer greater then or equal to $n$.
\end{abstract}

\keywords{Bernoulli polynomials, binomial coefficients, Bernoulli numbers}

\ams{11B68, 05A10, 05A15}

\maketitle

\section{\label{Sec1}Introduction}

Over the years, Bernoulli numbers $B_{n}$ and polynomials $B_{n}(x)$ have proven to be important mathematical objects. Since they
appeared on the scene (17th century), they have been of interest to many
mathematicians and they have appeared in many different fields of
mathematics (see the interesting document of Mazur \cite{Mazur}). Bernoulli
polynomials are usually defined by means of the generating function \cite[p.
48]{Comtet1974}%
\begin{eqnarray*}
\frac{te^{xt}}{e^{t}-1}=\sum_{n=0}^{\infty }B_{n}(x)\frac{t^{n}}{n!} \quad(|t|<2\pi).
\end{eqnarray*}

We can also find Bernoulli polynomials defined by \cite[p. 367]{GKP1994}:
\begin{eqnarray*}
B_{n}(x)=\sum_{k=0}^{n}\binom{n}{k}B_{k}x^{n-k}.
\end{eqnarray*}

We mention the explicit formula \cite[Vol. 8, (2.6)]{Gould2010}
\begin{equation}
B_{n}(x)=\sum_{i=0}^{n}\frac{1}{i+1}\sum_{j=0}^{i}\left(
-1\right) ^{j}\binom{i}{j}(x+j)^{n},
\label{1.3}
\end{equation}
since this is the known formula for Bernoulli polynomials that will be used
in this paper.

A large number of explicit formulas for Bernoulli numbers and polynomials
are known nowadays \cite{Costabile2006,Fekih-Ahmed,Vella2008}. We refer the reader to the article of
Gould \cite{Gould1972b} and mention, in particular, formula (3) of that article,
namely
\begin{equation}
B_{n}=\frac{1}{n+1}\sum_{i=1}^{n}\sum_{j=1}^{i}(-1)^{i+j}\frac{
\binom{n+1}{j}}{\binom{n}{i}}(i-j)^{n},
\label{1.4}
\end{equation}%
which is \textquotedblleft an old result\textquotedblright\ rediscovered in
1959. (Note that we can write (\ref{1.4}) with the indices $i,j$ beginning
at $0$. See also \cite[Vol. 8, (1.9)]{Gould2010} and \cite[(24.6.2)]{Dilcher}.)

The first main result we will prove in this paper is

\begin{proposition}
\label{Prop.1.1}For $0\leq n\leq k$ we have
\begin{equation}
(k+1)(-1)^{n}\sum_{i=0}^{n}\frac{1}{i+1}
\sum_{j=0}^{i}(-1)^{j}\binom{i}{j}(j+1)
^{n}=\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}
}{\binom{k}{i}}(j-i-1)^{n}.
\label{1.5}
\end{equation}
\end{proposition}

By writing (\ref{1.5}) as
\begin{equation}
(-1)^{n}\sum_{i=0}^{n}\frac{1}{i+1}\sum_{j=0}^{i}\left(
-1\right) ^{j}\binom{i}{j}(j+1)^{n}=\frac{1}{k+1}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{
\binom{k}{i}}(j-i-1)^{n},
\label{1.51}
\end{equation}
we can see that (\ref{1.5}) is an identity for the Bernoulli number $B_{n}$.
In fact, the left-hand side of (\ref{1.51}) is, according to (\ref{1.3}),
the Bernoulli number $B_{n}=(-1)^{n}B_{n}(1)$.
That is, formula (\ref{1.51}) says that $B_{n}$ is equal to the expression
of the right-hand side of (\ref{1.51}), \textit{for any} $k\geq n$.

The second main result of this paper says that formula (\ref{1.5}) is the
particular case $x=0$ of an identity involving Bernoulli polynomials.

\begin{proposition}
\label{Prop.1.2}For $0\leq n\leq k$ we have
\begin{equation}
B_{n}(x)=\frac{1}{k+1}\sum_{i=0}^{k}\sum_{j=0}^{i}\left(
-1\right) ^{i+j}\frac{\binom{k+1}{j}}{\binom{k}{i}}(x-i+j-1)^{n}.
\label{1.6}
\end{equation}
\end{proposition}

By using the known property of Bernoulli polynomials $(-1)^{n}B_{n}(x)=B_{n}(1-x)$, identity (\ref{1.6}) can
be written as%
\begin{equation}
B_{n}(x)=\frac{1}{k+1}\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{\binom{k}{i}}(x+i-j)^{n},
\label{1.7}
\end{equation}
and then identity (\ref{1.4}) is the particular case of (\ref{1.6}) in which
$x=0$ and $k=n$.

In the spirit of Gould's article \cite{Gould1972b} mentioned before, it is
difficult to claim originality for (\ref{1.6}). However, according to some
of the non-trivial consequences of this formula (we already began to explore
them \cite{Pita}), which are not widely known, we believe that it must have
some ingredient of originality.

In Section \ref{Sec2}, we state and prove four lemmas that will be used in
Section \ref{Sec3}, where we present the proofs of Propositions \ref%
{Prop.1.1} and \ref{Prop.1.2}. In Section \ref{Sec4}, we give some concrete
examples of (\ref{1.6}).

Throughout the paper, $k$ denotes a non-negative integer. We will be using,
without further comments, standard results about interchanging indices in
double sums (for example, those contained in Gould's book \cite[Vol. 1]{Gould2010}).

\section{\label{Sec2}Some lemmas}
Besides the basic facts involving binomial coefficients (e.g., the basic combinatorial identities),
we will need some more specialized results of this kind. More specifically, in the proofs of the
lemmas of this section, we will use five combinatorial identities from Gould's book \cite{Gould1972}.
For the reader's convenience, we quote them here:
\begin{eqnarray}
\sum_{k=0}^{n}(-1)^{k}\dbinom{x}{k}&=&(-1)^{n}\dbinom{x-1}{n}.
\label{1.10}
\\
\sum_{k=0}^{n}(-1)^{k}\dbinom{n}{k}k^{j}&=&
\begin{cases}
0, & \hbox{if $0\leq{j}<{n}$};\\
(-1)^n n!, & \hbox{if $j=n$}.
\end{cases}
\label{1.11}
\\
\sum_{k=1}^{n}\dbinom{x+k}{k}\dfrac{1}{x+k}&=&\dfrac{1}{x}\left(
\dbinom{x+n}{n}-1\right).
\label{1.12}
\\
\sum_{k=0}^{n}\dfrac{(-1) ^{k}}{\binom{x}{k}}&=&\dfrac{x+1%
}{x+2}\left(1+\dfrac{(-1)^{n}}{\binom{x+1}{n+1}}\right).
\label{1.13}
\\
\sum_{k=j}^{n}(-1)^{k}\dbinom{n}{k}\dbinom{k}{j}&=&
\begin{cases}
0, & \hbox{if $j\neq {n}$}; \\
(-1)^{n}, & \hbox{if $j=n$}.
\end{cases}
\label{1.14}
\end{eqnarray}

(In Gould's book \cite{Gould1972}, these are identities (1.5), (1.13), (1.50),
(2.1) and (3.119), respectively.)

\begin{lemma}
\label{Lem2.1}For $k\geq 0$ we have
\begin{equation}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{%
\binom{k}{i}}=k+1.
\label{2.1}
\end{equation}
\end{lemma}

\begin{proof}
Observe that
\begin{eqnarray*}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{
\binom{k}{i}} &=&\sum_{j=0}^{k}\sum_{i=0}^{j}(-1)^{i}\frac{
\binom{k+1}{k-j}}{\binom{k}{i+k-j}} \\
&=&\sum_{j=0}^{k}\sum_{i=0}^{j}(-1)^{i}\frac{\binom{k+1}{j+1}}{
\binom{k}{j-i}} \\
&=&\sum_{j=0}^{k}(-1)^{j}\binom{k+1}{j+1}\sum_{i=0}^{j}\left(
-1\right) ^{i}\frac{1}{\binom{k}{i}}.
\end{eqnarray*}

By using identity (\ref{1.13}) we obtain
\begin{eqnarray*}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{
\binom{k}{i}} &=&\sum_{j=0}^{k}(-1)^{j}\binom{k+1}{j+1}\frac{k+1
}{k+2}\left(1+\frac{(-1)^{j}}{\binom{k+1}{j+1}}\right) \\
&=&\frac{k+1}{k+2}\left(\sum_{j=0}^{k}(-1)^{j}\binom{k+1}{j+1}
+k+1\right) \\
&=&k+1\,,
\end{eqnarray*}
as desired.
\end{proof}

\begin{lemma}
\label{Lem2.2}
For $0\leq i\leq k$ and real $t$ with $t\neq 0$, we have
\begin{enumerate}
\item[$($a$)$]
\begin{equation}
1-t^{k+2}\sum_{j=0}^{i+1}\binom{k+2}{j}(t^{-1}-1)^{j}=(-1)^{i}(t-1)^{2+i}\sum_{j=0}^{k-i}\binom{j+i+1}{i+1}
t^{j}.
\label{2.2}
\end{equation}
\item[$($b$)$]
\begin{equation}
1-t^{k+1}\sum_{j=0}^{i}\binom{k+1}{j}(t^{-1}-1)^{j}=(1-t)^{1+i}\sum_{j=0}^{k-i}\binom{j+i}{i}t^{j}.
\label{2.3}
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
(a): We can write formula (\ref{2.2}) as
\begin{equation}
(1-t)^{2+i}\sum_{j=0}^{k-i}\binom{j+i+1}{i+1}
t^{j}+\sum_{j=0}^{i+1}\binom{k+2}{j}t^{k+2-j}\left(1-t\right)^{j}=1.
\label{2.4}
\end{equation}
Let us write $P(t)$ for the $(k+2)$-th degree
polynomial of the left-hand side in (\ref{2.4}). Observe that $P\left(
0\right) =1$, so we need to prove that the remaining coefficients of $P(t)$ are equal to $0$. We have
\begin{eqnarray}\label{2.5}
\begin{aligned}
P(t)=&\sum_{j=0}^{k-i}\sum_{r=0}^{i+2}\binom{k+1-j}{i+1}\binom{i+2}{r}(-1)^{r}t^{k+r-i-j}\\
&+\sum_{j=0}^{i+1}\sum_{r=0}^{j}\binom{k+2}{j}\binom{j}{r}(-1)^{r}t^{k+2+r-j}.
\end{aligned}
\end{eqnarray}
Let us introduce a new index $s$ in the sums of the left-hand side of (\ref{2.5}), namely $s=k+r-i-j$ for the first sum, and $s=k+2+r-j$ for the second
one. In both cases we have $0\leq s\leq k+2$, and we can write (\ref{2.5}) as
\begin{eqnarray*}
P(t)&=&(-1) ^{k}\sum_{s=0}^{k+2}\!\left(
\sum_{j=0}^{k-i}\binom{k+1-j}{i+1}\!\binom{i+2}{s+i+j-k}(-1)^{i+s+j}\right.\\
&& \left.+\sum_{j=0}^{i+1}\binom{k+2}{j}\!\binom{j}{s+j-k-2}(-1)^{s+j}\!\right) \!t^{s}
\end{eqnarray*}
or
\begin{eqnarray*}
P(t)&=&(-1)^{k}\sum_{s=0}^{k+2}\left(\sum_{j=0}^{k-i}\binom{i+1+j}{j}\binom{i+2}{s-j}(-1)^{s+j}\right.\\
&&\left.+\sum_{j=0}^{i+1}\binom{k+2}{j}\binom{j}{k+2-s}(-1)^{s+j}\right)t^{s}.
\end{eqnarray*}

We need to show that for $0\leq i\leq k$ and $1\leq s\leq k+2$ we have
\begin{equation}
\sum_{j=0}^{k-i}\binom{i+1+j}{j}\binom{i+2}{s-j}(-1)^{j}+(-1)^{k}\sum_{j=0}^{i+1}\binom{k+2}{j}\binom{j}{k+2-s}(-1)
^{j}=0.
\label{2.7}
\end{equation}

According to (\ref{1.14}), for $1\leq s\leq k+2$ we have
\begin{eqnarray*}
\sum_{j=0}^{k+2}\binom{k+2}{j}\binom{j}{k+2-s}(-1)^{j}=0\,,
\end{eqnarray*}
so we can write formula (\ref{2.7}) as
\begin{eqnarray*}
\sum_{j=0}^{k-i}\binom{i+1+j}{j}\binom{i+2}{s-j}(-1)^{j}-\left(
-1\right) ^{k}\sum_{j=i+2}^{k+2}\binom{k+2}{j}\binom{j}{k+2-s}\left(
-1\right) ^{j}=0
\end{eqnarray*}
or
\begin{equation}
\sum_{j=0}^{k-i}\binom{i+1+j}{j}\binom{i+2}{s-j}(-1)^{j}-\sum_{j=0}^{k-i}\binom{k+2}{j}\binom{k+2-j}{s-j}(-1)^{j}=0.
\label{2.8}
\end{equation}

Observe that
\begin{equation}
\sum_{j=0}^{k-i}\binom{k+2}{j}\binom{k+2-j}{s-j}(-1)^{j}=\left(
-1\right) ^{k+i}\binom{k+2}{s}\binom{s-1}{k-i},
\label{2.81}
\end{equation}
(identity (\ref{1.10}), together with some trivial simplifications).
We have also that
\begin{equation}
\sum_{j=0}^{k-i}\binom{i+1+j}{j}\binom{i+2}{s-j}(-1)^{j}=(-1)^{k+i}\binom{k+2}{s}\binom{s-1}{k-i}\,,
\label{2.82}
\end{equation}
which can be demonstrated by means of an induction argument on $k$ (left to
the reader). Identities (\ref{2.81}) and (\ref{2.82}) prove (\ref{2.8}), and
then the proof of (\ref{2.2}) is complete.

(b): The argument that proves (\ref{2.3}) is similar to that presented in
(a), and we show only some steps of this argument. We write (\ref{2.3}) as
\begin{eqnarray*}
(1-t)^{1+i}\sum_{j=0}^{k-i}\binom{j+i}{i}t^{j}+t^{k+1}
\sum_{j=0}^{i}\binom{k+1}{j}(t^{-1}-1)^{j}=1
\end{eqnarray*}
or
\begin{equation}
\sum_{j=0}^{k-i}\sum_{l=0}^{1+i}\binom{1+i}{l}\binom{j+i}{i}(-1)^{l}t^{l+j}+\sum_{j=0}^{i}\sum_{l=0}^{j}\binom{k+1}{j}\binom{j}{l}\left(
-1\right) ^{l}t^{k+1-j+l}=1.
\label{2.9}
\end{equation}

If $P(t)$ is the left-hand side polynomial in (\ref{2.9}), we
can see at once that $P(0)=1$, so we will prove that the
coefficient of $t^{s}$ in the polynomial $P(t)$ is equal to $0$
for $s\geq 1$. We set $s=l+j$ in the first sum of the left-hand side of (\ref
{2.9}), and we set $s=k+1-j+l\,$\ in the second one. Then we write (\ref{2.9}) as
\begin{eqnarray*}
&&\sum_{s=0}^{k+1}\sum_{j=0}^{i}\binom{k+1-i}{s-j}\binom{k+j-i}{j}\left(
-1\right) ^{s+j}t^{s}\\
&&\qquad+(-1)^{k}\sum_{s=i+1}^{k+1}\sum_{j=0}^{k-i}
\binom{k+1}{j}\binom{j}{k+1-s}(-1)^{s+j+1}t^{s}=1\,,
\end{eqnarray*}
so we need to show that for $0\leq i\leq k$ and $1\leq s\leq k+1$ we have
\begin{eqnarray}
\!\!\!\!\!\!\!\!\sum_{j=0}^{i}\binom{k+1-i}{s-j}\binom{k+j-i}{j}(-1)^{j}-\left(
-1\right) ^{k}\sum_{j=0}^{k-i}\binom{k+1}{j}\binom{j}{k+1-s}(-1)^{j}=0.
\label{2.11}
\end{eqnarray}

In fact, one can check that the identity
\begin{equation}
(-1)^{k}\sum_{j=0}^{k-i}\binom{k+1}{j}\binom{j}{k+1-s}\left(
-1\right) ^{j}=(-1)^{i}\binom{k+1}{s}\binom{s-1}{i}
\label{2.12}
\end{equation}
is equivalent to $\sum_{j=0}^{k-i}\binom{s}{j+i+1}(-1)^{j}=
\binom{s-1}{i}$, which in turn can be demonstrated easily. On the other
hand, we also have
\begin{equation}
\sum_{j=0}^{i}\binom{k+1-i}{s-j}\binom{k+j-i}{j}(-1)^{j}=(-1)^{i}\binom{k+1}{s}\binom{s-1}{i}\,,
\label{2.13}
\end{equation}
which can be written as
\begin{equation}
\sum_{j=0}^{k-i}\binom{i+1}{s-j}\binom{i+j}{j}(-1)^{j}=\left(
-1\right) ^{k+i}\binom{k+1}{s}\binom{s-1}{k-i}.
\label{2.14}
\end{equation}

Identity (\ref{2.14}) can be demonstrated with an induction argument on $k$
(left to the reader). Identities (\ref{2.12}) and (\ref{2.13}) prove (\ref
{2.11}), and then the proof of (\ref{2.3}) is complete.
\end{proof}

\begin{lemma}
\label{Lem2.3}
For $0\leq i\leq k$ we have
\begin{enumerate}
\item[$($a$)$]
\begin{equation}
\sum_{j=i}^{k}\frac{\binom{k+2}{j-i}}{\binom{k}{j}}=\frac{k+1}{i+1}\binom{k+1
}{i+1}.
\label{2.15}
\end{equation}
\item[$($b$)$]
\begin{equation}
\sum_{j=i}^{k}\frac{\binom{k+1}{j-i}}{\binom{k}{j}}=(k+1)
\sum_{j=0}^{k-i}\frac{1}{i+j+1}\binom{j+i}{i}.
\label{2.16}
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
We use that
\begin{equation}
\binom{n}{m}^{-1}=(n+1)\int_{0}^{1}t^{m}(1-t)^{n-m}dt\,,
\label{2.17}
\end{equation}
which is a popular tool used in problems where reciprocals of binomial
coefficients are involved \cite{Rockett1981,Sury2004,Trif2000,Yang2006}. The integral of the right-hand
side of (\ref{2.17}) is \\
$B\!\left(m+1,n-m+1\right)$, the Beta function
evaluated at $\left(m+1,n-m+1\right)$.

(a): We use (\ref{2.17}) to write the left-hand side of (\ref{2.15}) as
\begin{eqnarray*}
\sum_{j=i}^{k}\frac{\binom{k+2}{j-i}}{\binom{k}{j}}=\sum_{j=i}^{k}\binom{k+2
}{j-i}(k+1)\int_{0}^{1}t^{j}(1-t)^{k-j}dt.
\end{eqnarray*}
Then we have to show that
\begin{eqnarray*}
\int_{0}^{1}(1-t)^{k}\sum_{j=i}^{k}\binom{k+2}{j-i}\left(\frac{
t}{1-t}\right) ^{j}dt=\frac{1}{i+1}\binom{k+1}{i+1}
\end{eqnarray*}
or
\begin{equation}
\int_{0}^{1}(1-t)^{k}\left(\frac{t}{1-t}\right)^{i}\sum_{j=0}^{k-i}\binom{k+2}{j}\left(\frac{t}{1-t}\right) ^{j}dt=\frac{1
}{i+1}\binom{k+1}{i+1}.  \label{2.19}
\end{equation}
Observe that
\begin{eqnarray}
\sum_{j=0}^{k-i}\binom{k+2}{j}\left(\frac{t}{1-t}\right)^{j}
&=&\sum_{j=0}^{k+2}\binom{k+2}{j}\left(\frac{t}{1-t}\right)
^{j}-\sum_{j=k-i+1}^{k+2}\binom{k+2}{j}\left(\frac{t}{1-t}\right)^{j}
\notag \\
&=&\left(\frac{t}{1-t}+1\right)^{k+2}-\sum_{j=0}^{i+1}\binom{k+2}{j}\left(
\frac{t}{1-t}\right) ^{k+2-j}  \notag \\
&=&\left(\frac{1}{1-t}\right)^{k+2}\left(1-t^{k+2}\sum_{j=0}^{i+1}\binom{
k+2}{j}\left(\frac{1-t}{t}\right)^{j}\right).
\label{2.191}
\end{eqnarray}
Use (\ref{2.191}) to write the left-hand side of (\ref{2.19}) as
\begin{eqnarray}\label{2.20}
\begin{aligned}
&\int_{0}^{1}(1-t)^{k}\left(\frac{t}{1-t}\right)^{i}\sum_{j=0}^{k-i}\binom{k+2}{j}\left(\frac{t}{1-t}\right)^{j}dt
\\
&\qquad=\int_{0}^{1}\left(\frac{1}{1-t}\right)^{2}\left(\frac{t}{1-t}\right)^{i}\left( 1-t^{k+2}\sum_{j=0}^{i+1}\binom{k+2}{j}(t^{-1}-1)^{j}\right) dt.
\end{aligned}
\end{eqnarray}
According to (\ref{2.2}), from (\ref{2.20}) we have that
\begin{eqnarray*}
&&\int_{0}^{1}(1-t)^{k}\left(\frac{t}{1-t}\right)^{i}\sum_{j=0}^{k-i}\binom{k+2}{j}\left(\frac{t}{1-t}\right)^{j}dt \\
&&\qquad=\int_{0}^{1}\left(\frac{1}{1-t}\right)^{2}\left(\frac{t}{1-t}\right)^{i}(-1)^{i}(t-1)^{2+i}\sum_{j=0}^{k-i}\binom{j+i+1
}{i+1}t^{j}dt.
\end{eqnarray*}

That is, we have that
\begin{eqnarray*}
\int_{0}^{1}(1-t)^{k}\left(\frac{t}{1-t}\right)^{i}\sum_{j=0}^{k-i}\binom{k+2}{j}\left(\frac{t}{1-t}\right)^{j}dt
&=&\sum_{j=0}^{k-i}\binom{j+i+1}{i+1}\int_{0}^{1}t^{i+j}dt \\
&=&\sum_{j=0}^{k-i}\frac{1}{i+j+1}\binom{j+i+1}{i+1} \\
&=&\frac{1}{i+1}\binom{k+1}{i+1}\,,
\end{eqnarray*}
as desired. (In the last step, we used (\ref{1.12}).)

(b): The left-hand side of (\ref{2.16}) can be written (according to (\ref
{2.17})) as
\begin{eqnarray*}
\sum_{j=i}^{k}\frac{\binom{k+1}{j-i}}{\binom{k}{j}}=(k+1)\sum_{j=i}^{k}\binom{k+1}{j-i}\int_{0}^{1}t^{j}(1-t)^{k-j}dt\,,
\end{eqnarray*}
and then we have to show that
\begin{eqnarray*}
\int_{0}^{1}(1-t)^{k}\sum_{j=i}^{k}\binom{k+1}{j-i}\left(\frac{
t}{1-t}\right) ^{j}dt=\sum_{j=0}^{k-i}\frac{1}{i+j+1}\binom{j+i}{i}
\end{eqnarray*}
or
\begin{equation}
\int_{0}^{1}(1-t)^{k}\left(\frac{t}{1-t}\right)^{i}\sum_{j=0}^{k-i}\binom{k+1}{j}\left(\frac{t}{1-t}\right)^{j}dt=\sum_{j=0}^{k-i}\frac{1}{i+j+1}\binom{j+i}{i}.
\label{2.22}
\end{equation}

As we did with (\ref{2.191}), we can see that
\begin{equation}
\sum_{j=0}^{k-i}\binom{k+1}{j}\left(\frac{t}{1-t}\right)^{j}=\left(\frac{1
}{1-t}\right)^{k+1}\left(1-t^{k+1}\sum_{j=0}^{i}\binom{k+1}{j}\left(\frac{
1-t}{t}\right)^{j}\right).
\label{2.23}
\end{equation}

Then, beginning with the left-hand side of (\ref{2.22}), using (\ref{2.23})
and (\ref{2.3}), we have that
\begin{eqnarray*}
&&\int_{0}^{1}(1-t)^{k}\left(\frac{t}{1-t}\right)^{i}\sum_{j=0}^{k-i}\binom{k+1}{j}\left(\frac{t}{1-t}\right)^{j}dt \\
&&\qquad=\int_{0}^{1}\left(\frac{1}{1-t}\right) \left(\frac{t}{1-t}\right)^{i}\left(1-t^{k+1}\sum_{j=0}^{i}\binom{k+1}{j}(t^{-1}-1)^{j}\right) dt \\
&&\qquad=\int_{0}^{1}\left(\frac{1}{1-t}\right) \left(\frac{t}{1-t}\right)^{i}(1-t)^{1+i}\sum_{j=0}^{k-i}\binom{j+i}{i}t^{j}dt \\
&&\qquad=\sum_{j=0}^{k-i}\binom{j+i}{i}\int_{0}^{1}t^{i+j}dt \\
&&\qquad=\sum_{j=0}^{k-i}\frac{1}{i+j+1}\binom{j+i}{i}\,,
\end{eqnarray*}
as desired.
\end{proof}

\begin{lemma}
\label{Lem2.4}
For $2\leq m\leq k$ we have
\begin{equation}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+2}{j}}{
\binom{k}{i}}(j-i-1)^{m}=0.
\label{2.24}
\end{equation}
\end{lemma}

\begin{proof}
We write the left-hand side of (\ref{2.24}) as
\begin{eqnarray*}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+2}{j}}{
\binom{k}{i}}(j-i-1)^{m}=(-1)^{m}\sum_{i=0}^{k}(-1)^{i}(i+1)^{m}\sum_{j=i}^{k}
\frac{\binom{k+2}{j-i}}{\binom{k}{j}}.
\end{eqnarray*}

Thus, by using (\ref{2.15}) (and the fact that $m\geq 2$) we have that
\begin{eqnarray}
&&\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+2}{j}}{
\binom{k}{i}}(j-i-1)^{m}  \notag \\
&&\qquad=(-1)^{m}\sum_{i=0}^{k}(-1)^{i}(i+1)^{m}\frac{k+1}{i+1}\binom{k+1}{k-i}  \notag \\
&&\qquad=(-1)^{m+k}(k+1) \sum_{i=0}^{k}(-1)^{i}\frac{(k+1-i)^{m}}{k+1-i}\binom{k+1}{i}  \notag \\
&&\qquad=(-1)^{m+k}(k+1) \sum_{i=0}^{k+1}\binom{k+1}{i}
(-1)^{i}(k+1-i)^{m-1}  \notag \\
&&\qquad=(-1)^{m+k}(k+1) \!\!\!\sum_{l=0}^{m-1}(-1)^{l}\binom{m-1}{l}(k+1)^{m-1-l}\!\!\sum_{i=0}^{k+1}\binom{k+1}{i}
(-1)^{i}i^{l}.
\label{2.245}
\end{eqnarray}

Identity (\ref{1.11}) tells us that for $0\leq l\leq k$ we have
\begin{equation}
\sum_{i=0}^{k+1}\binom{k+1}{i}(-1)^{i}i^{l}=0.
\label{2.242}
\end{equation}

In particular, for $0\leq l\leq m-1$, identity (\ref{2.242}) is valid, and
then (\ref{2.245}) is equal to $0$, finishing the proof of (\ref{2.24}).
\end{proof}


\section{\label{Sec3}Proofs of Propositions \protect\ref{Prop.1.1} and
\protect\ref{Prop.1.2}}

Now we are ready to prove the main results of the article.

\begin{proof}[\bf Proof of Proposition \ref{Prop.1.1}.]

Beginning with the right-hand side of (\ref{1.5}), we can write
\begin{eqnarray*}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{
\binom{k}{i}}(j-i-1)^{n}=(-1)^{n}\sum_{i=0}^{k}(-1)^{i}(i+1)^{n}\sum_{j=i}^{k}
\frac{\binom{k+1}{j-i}}{\binom{k}{j}}.
\end{eqnarray*}
According to (\ref{2.16}), we have
\begin{eqnarray*}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{
\binom{k}{i}}(j-i-1)^{n}&=&(k+1) (-1)^{n}\sum_{i=0}^{k}(-1)^{i}(i+1)^{n}\\
&&\times\sum_{j=0}^{k-i}\frac{1}{i+j+1}\binom{j+i}{i}
\end{eqnarray*}
or
\begin{eqnarray*}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{
\binom{k}{i}}(j-i-1)^{n}=(k+1)(-1)^{n}\sum_{i=0}^{k}(-1)^{i}(i+1)^{n}\sum_{j=i}^{k}
\frac{1}{j+1}\binom{j}{i}.
\end{eqnarray*}

Some further simplifications give us
\begin{eqnarray*}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{
\binom{k}{i}}(j-i-1)^{n}&=&(k+1) (-1)^{n}\sum_{j=0}^{k}(-1)^{j}(j+1)^{n}\sum_{i=j}^{k}
\frac{1}{i+1}\binom{i}{j} \\
&=&(k+1) (-1)^{n}\sum_{i=0}^{k}\frac{1}{i+1}
\sum_{j=0}^{k}(-1)^{j}\binom{i}{j}(j+1)^{n}.
\end{eqnarray*}
Finally, we claim that for $0\leq n\leq k$ we have
\begin{equation}
\sum_{i=0}^{k}\frac{1}{i+1}\sum_{j=0}^{k}(-1)^{j}\binom{i}{j}
(j+1)^{n}=\sum_{i=0}^{n}\frac{1}{i+1}\sum_{j=0}^{i}\left(
-1\right) ^{j}\binom{i}{j}(j+1)^{n}.
\label{2.29}
\end{equation}

In fact, beginning with the left-hand side of (\ref{2.29}) we have that
\begin{eqnarray*}
\sum_{i=0}^{k}\frac{1}{i+1}\sum_{j=0}^{k}(-1)^{j}\binom{i}{j}
(j+1)^{n}&=&\sum_{i=0}^{n}\frac{1}{i+1}\sum_{j=0}^{i}\left(
-1\right) ^{j}\binom{i}{j}(j+1)^{n}\\
&&+\sum_{i=n+1}^{k}\frac{1}{i+1}\sum_{j=0}^{i}(-1)^{j}\binom{i}{j}(j+1)^{n}.
\end{eqnarray*}

But for $i\geq n+1$ we have $\sum_{j=0}^{i}(-1)^{j}\binom{i}{j}
(j+1)^{n}=0$ (identity (\ref{1.11})), from where our claim (\ref
{2.29}) follows, and the proof is complete.
\end{proof}

\begin{proof}[\bf Proof of Proposition \ref{Prop.1.2}.]
Let us denote the right-hand side polynomial in (\ref{1.6}) as $\mathbb{B}
_{n}(x)$. That is,
\begin{equation}
\mathbb{B}_{n}(x)=\frac{1}{k+1}\sum_{i=0}^{k}\sum_{j=0}^{i}
(-1)^{i+j}\frac{\binom{k+1}{j}}{\binom{k}{i}}\left(
x-i+j-1\right) ^{n}.
\label{3.2}
\end{equation}
The proof is based on the following characterization of Bernoulli
polynomials \cite[pp. 17--18]{Norlund1924}: \textit{the only polynomial }$\mathbb{B}
_{n}(x)$\textit{\ such that }$\mathbb{B}_{n}(x+1) -
\mathbb{B}_{n}(x)=nx^{n-1}$\textit{, }$\mathbb{B}_{n}\left(
0\right) =B_{n}$\textit{, is the }$n$\textit{-th Bernoulli polynomial }$
B_{n}(x)$. Proposition \ref{Prop.1.1} tells us that $\mathbb{B}
_{n}(0)=B_{n}$. Thus, we need to prove that the polynomial (\ref
{3.2}) satisfies the difference equation $\mathbb{B}_{n}(x+1) -
\mathbb{B}_{n}(x)=nx^{n-1}$. That is, we need to prove that
\begin{eqnarray*}
\frac{1}{k+1}\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{
k+1}{j}}{\binom{k}{i}}\left((x-i+j)^{n}-(x-i+j-1)^{n}\right)=nx^{n-1}.
\end{eqnarray*}
Observe that
\begin{eqnarray*}
&&\frac{1}{k+1}\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{
\binom{k+1}{j}}{\binom{k}{i}}\left((x-i+j)^{n}-(x-i+j-1)^{n}\right) \\
&&\qquad=\sum_{r=0}^{n}\binom{n}{r}\frac{1}{k+1}\sum_{i=0}^{k}\sum_{j=0}^{i}\left(
-1\right) ^{i+j}\frac{\binom{k+1}{j}}{\binom{k}{i}}\left((j-i)^{n-r}-(j-i-1)^{n-r}\right) x^{r} \\
&&\qquad=nx^{n-1}\!+\!\!\sum_{r=0}^{n-2}\binom{n}{r}\frac{1}{k+1}\!\!\sum_{i=0}^{k}
\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{\binom{k}{i}}
\left((j-i)^{n-r}-(j-i-1)^{n-r}\right) x^{r}
\end{eqnarray*}
(in the last step, we used (\ref{2.1})). Thus, our proof ends if we show that
for $0\leq r\leq n-2$ we have
\begin{equation}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{
\binom{k}{i}}\left((j-i)^{n-r}-(j-i-1)^{n-r}\right) =0.
\label{3.4}
\end{equation}
We will prove that for $2\leq m\leq k$ we have the identity
\begin{equation}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{
\binom{k}{i}}(j-i)^{m}=\sum_{i=0}^{k}\sum_{j=0}^{i}\left(
-1\right) ^{i+j}\frac{\binom{k+1}{j}}{\binom{k}{i}}(j-i-1)^{m}\,,
\label{3.5}
\end{equation}
which plainly implies that we have (\ref{3.4}) for $0\leq r\leq n-2$, and
which in turn gives us the desired end of the proof.

We can write (\ref{3.5}) as
\begin{equation}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j+1}\frac{\binom{k+1}{j-1}}{
\binom{k}{i}}(j-i-1)^{m}=\sum_{i=0}^{k}\sum_{j=0}^{i}\left(
-1\right) ^{i+j}\frac{\binom{k+1}{j}}{\binom{k}{i}}(j-i-1)^{m}\, ,
\label{3.6}
\end{equation}%
and we can easily see that (\ref{3.6}) is equivalent to
\begin{equation}
\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+2}{j}}{
\binom{k}{i}}(j-i-1)^{m}=0.
\label{3.7}
\end{equation}

But in Lemma \ref{Lem2.4} we proved that (\ref{3.7}) is true for $2\leq
m\leq k$. Then our proof is complete.
\end{proof}



\section{\label{Sec4}Examples}

In Proposition \ref{Prop.1.2}, we demonstrated that the Bernoulli polynomial $
B_{n}(x)$ can be written as a linear combination of $(k+1) $ polynomials
$( x-1)^{n}$,$(x-2)^{n}$, $\ldots$, $(x-k-1)^{n}$, where $k$ is any positive integer $\geq
n$. An additional observation is that Lemma \ref{Lem2.1} says that this
linear combination is in fact an \textit{affine} combination: if
\[
B_{n}(x)=\sum_{t=0}^{k}c_{t}(x-1-t)^{n},
\]
the
coefficients $c_{0},\ldots ,c_{k}$ satisfy that
\[
\sum_{t=0}^{k}c_{t}=1.
\]
In this section, we give some concrete examples.

With $k=1$ formula (\ref{1.6}) looks as
\begin{equation}
B_{n}(x)=\frac{3}{2}(x-1)^{n}-\frac{1}{2}(x-2)^{n}\,,
\label{4.1}
\end{equation}
which works for $n=0,1$, namely
\begin{eqnarray*}
B_{0}(x)&=&\frac{3}{2}(x-1)^{0}-\frac{1}{2}(x-2)^{0}=1, \\
B_{1}(x)&=&\frac{3}{2}(x-1)-\frac{1}{2}(x-2)=x-\frac{1}{2}.
\end{eqnarray*}

(For \thinspace $n>1$, the right-hand side of (\ref{4.1}) does not give us $
B_{n}(x)$. For example, if $n=2$, the right-hand side of (\ref
{4.1}) is $\frac{3}{2}(x-1)^{2}-\frac{1}{2}(x-2)^{2}=\allowbreak x^{2}-x-\frac{1}{2}$,
which is not $B_{2}(x)=x^{2}-x+\frac{1}{6}$.)

With $k=2$ formula (\ref{1.6}) is
\begin{eqnarray*}
B_{n}(x)=\frac{11}{6}(x-1)^{n}-\frac{7}{6}(x-2)^{n}+\frac{1}{3}(x-3)^{n}\,,
\end{eqnarray*}
which works for $n=0,1,2$, namely
\begin{eqnarray*}
B_{0}(x)&=&\frac{11}{6}(x-1)^{0}-\frac{7}{6}(x-2)^{0}+\frac{1}{3}(x-3)^{0}=1, \\
B_{1}(x)&=&\frac{11}{6}(x-1)-\frac{7}{6}(x-2)+\frac{1}{3}(x-3)=x-\frac{1}{2}, \\
B_{2}(x)&=&\frac{11}{6}(x-1) ^{2}-\frac{7}{6}(x-2)^{2}+\frac{1}{3}(x-3)^{2}=x^{2}-x+\frac{1}{6}.
\end{eqnarray*}

From a different point of view, for each
non-negative integer $n$,  expression (\ref{1.6}) gives us infinitely many formulas to write the $n$-th
Bernoulli polynomial $B_{n}(x)$, namely, formulas of the
right-hand side of (\ref{1.6}) with $k\geq n$. For example, the first 3
formulas for the Bernoulli polynomial $B_{3}(x)=x^{3}-\frac{3}{2
}x^{2}+\frac{1}{2}x$ are the following:
\begin{eqnarray*}
B_{3}(x)&=&\frac{25}{12}(x-1)^{3}-\frac{23}{12}
(x-2)^{3}+\frac{13}{12}(x-3)^{3}-\frac{1}{4}(x-4)^{3}, \\
&=&\frac{137}{60}(x-1)^{3}-\frac{163}{60}(x-2)^{3}+\frac{137}{60}(x-3)^{3}-\frac{21}{20}(x-4)^{3}+
\frac{1}{5}(x-5)^{3}, \\
&=&\frac{49}{20}(x-1)^{3}-\frac{71}{20}(x-2)^{3}+\frac{79}{20}(x-3)^{3}\\
&&-\frac{163}{60}(x-4)^{3}+\frac{31}{30}(x-5)^{3}-\frac{1}{6}(x-6)^{3}
\end{eqnarray*}
(corresponding to $k=3,4,5$). In particular, the Bernoulli polynomial $
B_{0}(x)=1$ can be written as
\begin{eqnarray*}
B_{0}(x)=\frac{1}{k+1}\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{\binom{k}{i}}\,,
\end{eqnarray*}%
where $k$ is any non-negative integer. (This affirmation is precisely (\ref
{2.1}).) Similarly, the Bernoulli polynomial $B_{1}(x)=x-\frac{1
}{2}$ can be written as
\begin{eqnarray*}
B_{1}(x)=\frac{1}{k+1}\sum_{i=0}^{k}\sum_{j=0}^{i}(-1)^{i+j}\frac{\binom{k+1}{j}}{\binom{k}{i}}(x-i+j-1)\,,
\end{eqnarray*}
where $k$ is any positive integer.



%%%% Acknowledgment %%%%%%%%
\section*{Acknowledgement}
The authors would like to thank the referee for the helpful suggestions.
%This work has been partly done when the first author visited Universidad Panamericana in 2005.
%He thanks the warm hospitality for the second author and his institute.
The first author is supported in part by the Hubei Provincial Expert Program and the grant by Wuhan University.



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