The coefficients of a series related to the Bernoulli polynomials 
The generating function for Bernoulli polynomials is given by:
  $$\frac{ue^{ux}}{e^u-1}=\sum_{n\geq 0}B_n(x)\frac{u^n}{n!}$$  

Now, I have the following expression:
$$\frac{1}{\alpha}\frac{u^2e^{u(\alpha x+y)}}{e^u-1}\sum_{k=1}^{\alpha-1}\frac{1}{e^u-e^{2\pi ik/\alpha}}$$
and I want to rewrite it in terms of Bernoulli polynomials and then extract the coefficients of $\frac{u^n}{n!}$. 
 A: With the recent edit something like the following is possible but it might not be as pretty as you were hoping; I'll put it up in case someone can spot how to improve it.
I presume $\alpha\in \mathbb{Z}$ as it appears as the upper limit in the sum, but if not the following will not hold. I also restrict $0< u$ so I can use a geometric series.
At a glance this looks like the product of two terms resembling the generating function, so we begin by saying your expression is the product of:
$${\frac {u{{\rm e}^{u \left( \alpha\,x+y \right) }}}{-1+{{\rm e}^{u}}}}
=\sum _{k=0}^{\infty }{\frac {B_k \left( \alpha\,x+y
 \right) {u}^{k}}{k!}}$$
and:
$$\frac{1}{\alpha}u\sum_{k=1}^{\alpha-1}\frac{1}{e^u-e^{2\pi ik/\alpha}}=u\frac{1}{\alpha}\sum _{k=1}^{\alpha-1}  \sum _{n=1}^{\infty }{{\rm e}^{-un}}{
{\rm e}^{{\frac {2\,i\pi \,k ( n-1 ) }{\alpha}}}} =u\sum _{n=1}^{\infty }{{\rm e}^{-un}}\frac{1}{\alpha}\sum _{k=1}^{\alpha-1}{
{\rm e}^{{\frac {2\,i\pi \,k ( n-1 ) }{\alpha}}}} 
 $$
where in the first line we simply used the formula and in the second we used a geometric series and then reversed the order of summation as the inner sum is absolutely convergent for $u>0$. Then note that:
$$\frac{1}{\alpha}\sum _{k=1}^{\alpha-1}{
{\rm e}^{{\frac {2\,i\pi \,k ( n-1 ) }{\alpha}}}}=\delta_{\alpha\, l,\,n-1}-\frac{1}{\alpha}=\delta_{\alpha\, l+1,\,n}-\frac{1}{\alpha}$$
where I've introduced $l$ to mean that the Kronecker delta will be $1$ only when $n-1$ is a multiple of $\alpha$ and $0$ otherwise; note also that because $n$ begins at $1$, $l$ begins at $0$. Taking this into account we have:
$$u\sum _{n=1}^{\infty }{{\rm e}^{-un}}\frac{1}{\alpha}\sum _{k=1}^{\alpha-1}{
{\rm e}^{{\frac {2\,i\pi \,k ( n-1 ) }{\alpha}}}} =u\sum _{l=0}^{\infty }{{\rm e}^{-u(\alpha\,l+1)}}-u\frac{1}{\alpha}\sum _{n=1}^{\infty }{{\rm e}^{-u\,n}}$$ $$=-{\frac {{{\rm e}^{-u}}u}{-1+{{\rm e}^{-u\alpha}}}}-{\frac {u}{
\alpha\, \left( -1+{{\rm e}^{u}} \right) }}=\frac{1}{\alpha}\sum _{j=0}^{\infty }{\frac { \left( B_j \left(\frac{1}{\alpha} \right)  \left( -\alpha \right) ^{j}-B_j \left( 0
 \right)  \right) {u}^{j}}{j!}}
$$
where we again summed the geometric series and then used the formula; note $B_j(0)$ is a Bernoulli number. You then have that:
$$\frac{1}{\alpha}\frac{u^2e^{u(\alpha x+y)}}{e^u-1}\sum_{k=1}^{\alpha-1}\frac{1}{e^u-e^{2\pi ik/\alpha}}=\left(\sum _{k=0}^{\infty }{\frac {B_k \left( \alpha\,x+y
 \right) {u}^{k}}{k!}}\right)\left( \frac{1}{\alpha}\sum _{j=0}^{\infty }{\frac { \left( B_j \left(\frac{1}{\alpha} \right)  \left( -\alpha \right) ^{j}-B_j \left( 0
 \right)  \right) {u}^{j}}{j!}} \right)$$
and after taking the Cauchy product of the two series:
$$\frac{1}{\alpha}\frac{u^2e^{u(\alpha x+y)}}{e^u-1}\sum_{k=1}^{\alpha-1}\frac{1}{e^u-e^{2\pi ik/\alpha}}=\sum_{n=0}^{\infty}c_n\frac{u^n}{n!}$$
where $c_n$ is the convolution:
$$c_n=\frac{1}{\alpha} \sum _{m=0}^{n}{\frac { n!\left( B_{{n-m}} \left( \frac{1}{\alpha} \right) 
 \left( -\alpha \right) ^{n-m}-B_{{n-m}} \left( 0 \right)  \right) B_{
{m}} \left( \alpha\,x+y \right) }{ \left( n-m \right) !\,m!}}$$
$$c_n=\frac{1}{\alpha} \sum _{m=0}^{n}{ { \binom{n}{m}\left( B_{{n-m}} \left( \frac{1}{\alpha} \right) 
 \left( -\alpha \right) ^{n-m}-B_{{n-m}} \left( 0 \right)  \right) B_{
{m}} \left( \alpha\,x+y \right) }} $$
When the series converges, this expression checks out numerically for the values I tested . I confess I have not found the upper bound on $u$ yet but I conjecture the series holds for $0< u<1$ although it could possibly have a dependence on $\alpha$ which I have not found yet.
