Hurwitz Zeta in terms of Bernoulli polynomials. @Raymond Manzoni showed nicely in 
this post how the Riemann zeta function is related to the Bernoulli numbers using the Euler-Maclaurin sum. The result is :
\begin{eqnarray}
\zeta(1-k) = -\frac{B_k}{k}.
\end{eqnarray}
(see his equation (4) )
Following a similar process but this time, instead of the Riemann Zeta function we use the Hurwitz zeta function $\zeta(s)=\sum_{n=0}^{\infty} 1/(n+x)^s$ , $\mathrm{Re}(x)>0$, we should find the extension of the equation above:
\begin{eqnarray}
\zeta(1-k,x) = -\frac{B_k(x)}{k}.
\end{eqnarray}
I am having difficulties with this. Any help is appreciated.
Thanks.

I found an answer:
We will need the following easy to prove (I can expand here if necessary)
relations:
\begin{eqnarray*}
 B_{k+1}(x) =  \sum_{i=0}^{k+1} B_i 
  \binom{k+1}{i}
   x^{k+1-i}   \quad \quad (1),
\end{eqnarray*}
and
\begin{eqnarray*}
  \sum_{i=0}^{n-1} (i+m)^{k-1} = 
  \frac{B_{k}(n+m) - B_{k}(m)}{k} \quad \quad (2)
\end{eqnarray*}
Let us start:
\begin{eqnarray*}
  \sum_{k=0}^{\infty} \frac{1}{(k+x)^s} = 
  \sum_{k=0}^{N-1} \frac{1}{(k+x)^s} + 
  \sum_{k=N}^{\infty} \frac{1}{(k+x)^s} 
\end{eqnarray*}
and, in the second sum, use the Euler-Maclaurin series with
$h=1, a=N, n= \infty$, $f(k)=1/(k+x)^s$. Then
\begin{eqnarray*} 
  \sum_{i=0}^{n-1} f(a + i h) h = \int_a^b f(t) dt + 
     \left . \sum_{k=1}^{m} \frac{B_k
       h^k}{k!} f^{(k-1)} (t)  \right |_a^b + R_m \\
   \sum_{i=0}^{\infty} \frac 1{(N + i+x)^s} = 
   \int_{N}^{\infty} \frac {dt}{(t+x)^s}  + \left . 
    \sum_{k=1}^{m} \frac{B_k}{k!} (-1)^{k-1}\frac{\Gamma(s+k-1)}{\Gamma(s) (t+x)^{s+k-1}}
    \right |_{N}^{\infty}+ R_m\\ 
    \sum_{k={N}}^{\infty} \frac 1{(k+x)^s} = \frac {1}{(s-1)(N+x)^{s-1}}  - 
    \sum_{k=1}^{m} \frac{B_k}{k!} (-1)^{k-1}\frac{\Gamma(s+k-1)}{\Gamma(s) \, 
       (N+x)^{s+k-1}}+ R_m
\end{eqnarray*}
with 
\begin{eqnarray*}
  R_m =  \int_N^{\infty} B_m \left \{ \frac{t-a}{h} \right \}
    \frac{\Gamma(s+m)}{ \Gamma(s)} \frac{1}{ (t+x)^{s+m+1}}.
\end{eqnarray*}
Then
\begin{eqnarray*}
 \zeta(s,x)= \sum_{k=0}^{N-1} \frac 1{(k+x)^s} +
 \frac {1}{(s-1)\,(N+x)^{s-1}} -
 \sum_{k=1}^{m} \frac{B_k}{k!} \frac{\Gamma(s+k-1)}{\Gamma(s) (N+x)^{s+k-1}}+ R_m,
\end{eqnarray*}
and taking the limit as $m \to \infty$ (since the
sum converges and $R_m \to 0$, this is easy to show for this function), we find
\begin{eqnarray*}
\zeta(s,x) =  \sum_{k=0}^{N-1} \frac{1}{(k+x)^s}  +
\frac{1}{ (s-1) (N+x)^{s-1}} + \sum_{k=1}^{\infty} \frac{B_k }{k!} 
\frac{\Gamma(s+k-1)}{\Gamma(s)}.
\end{eqnarray*}
We want to  choose $s=1-j$, for $j$ a positive integer.  First,
we consider the 
Pochhammer Symbold
\begin{eqnarray*}
  (s)_{k-1} = \frac{ \Gamma(s + k -1)}{\Gamma(s)}.
\end{eqnarray*}
We prove the reflection formula for the Pochhammer symbol,
\begin{eqnarray*}
(-t)_n = (-1)^n (t -n + 1)_n
\end{eqnarray*}
This is,
\begin{eqnarray*}
  (-t)_n &=& \frac{ \Gamma(-t+n)}{\Gamma(-t)} \\
  &=& \frac{ (-t+n-1)(-t+n-2) \cdots (-t) \cancel{\Gamma(-t)}}{\cancel{\Gamma(-t)}} \\
  &=& (-1)^n (t-n+1)(t-n+2) \cdots t \\
  &=& (-1)^n t (t-1) \cdots (t-n+1) \\
  &=& (-1)^n \frac{\Gamma(t+1)}{\Gamma(t-n+1)} \\
  &=& (-1)^n (t-n+1)_n
\end{eqnarray*}
Then we find that for $s=1-j$
\begin{eqnarray*}
  \frac{\Gamma(s+k-1)}{k! \; \Gamma(s)} = \frac{(1-j)_{k-1}}{k!} =
  (-1)^{k-1} \frac{(j-1-k+1 +1)_{k-1}}{k!}
  = (-1)^{k-1} \frac{ (j-k+1)_{k-1}}{k!} 
  = (-1)^{k-1} \frac{ (j-1)!}{k! \, \Gamma(j-k)}
\end{eqnarray*}
and we can write the formula for $\zeta(1-j,x)$ as
\begin{eqnarray*}
  \zeta(1-j,x) &=& 
  \sum_{k=0}^{N-1} (k+x)^{j-1} 
  - \frac{ (N+x)^j}{j} 
  + \frac{1}{j} \sum_{k=1}^j (-1)^{k-1} B_k  \;
  (N+x)^{j-k} \binom{j}{k} .
  \\
  &=&
  \sum_{k=0}^{N-1} (k+x)^{j-1} 
  - \frac{1}{j} \sum_{k=0}^j (-1)^{k} B_k  \;
  (N+x)^{j-k} \binom{j}{k} .
\end{eqnarray*}
where we introduced the independent term into the second sum. We will change 
the alternating sign $(-1)^{k-1}$ for a $-1$ sign,  since for $k > 1$, odd $B_{k}=0$, and the
first two terms (indices $k=0,1$) have the right sign.
We recognize, from equation (1), into the equation above that
\begin{eqnarray*}
  \zeta(1-j,x) = 
  \sum_{k=0}^{N-1} (k+x)^{j-1} 
  - \frac{1}{j} \, B_{j}(N+x).
\end{eqnarray*}
where we removed the alternating sign $(-1)^{k-1}$ since for $i>1$ odd $B_k=0$.
We now use equation (2) to write
\begin{eqnarray*}
  \zeta(1-j,x) = 
  \frac{B_{j}(N+x) - B_{j}(x)}{j}
  - \frac{1}{j} \, B_{j}(N+x)
  = -\frac{B_j(x)}{j}.
\end{eqnarray*}
This is what we wanted to show.
