The generating function of the Bernoulli Polynomials is: $$\frac{t e^{xt}}{e^t-1}=\sum\limits_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}.$$ Put $b_n=B_n(1)$. Consider that $B_n(1)=B_n(0)$ for $n \geq 2$ . Now let $$S_n(z)=\frac{B_{n+1}(z+1)-B_{n+1}(1)}{n+1}.$$ Note that if $m \in \mathbb{Z}^+$, then $S_n(m)=\sum^{m}_{k=1}k^n.$
Also we compute $\sum^{N}_{n=0}S_n(z)$ as below $$\sum^{N}_{n=0}S_n(z)=\sum^{N}_{n=0}\sum^{n+1}_{k=1}\frac{n!}{k!(n+1-k)!}b_{n+1-k}z^k=\sum^{N+1}_{n=1}\sum^{N}_{k=n-1}\frac{k!}{n!(k+1-n)!}b_{k+1-n}z^n.$$The last equality is taken from the following identity for iterated series of double complex sequences$$\sum^{N}_{n=0}\sum^{n+1}_{k=1}C_{nk}=\sum^{N+1}_{n=1}\sum^{N}_{k=n-1}C_{kn}.$$ Now I Investigate convergence of the sequence $\sum^{N}_{n=0}S_n(z)$ when $N\rightarrow \infty.$ Is it convergent for every $z?$ If not, for which $z$ is it convergent?