Sum with Bernoulli polynomial

The last sum presented won't converge. Here is a simpler approach: $$\sum_{k=0}^n \binom{n}{k}\frac{B_k(x)}{n-k+1} = \sum_{k=0}^n \binom{n}{k}B_k(x) \int_0^1 u^{n-k} du = \int_0^1 B_n(x+u) du$$ where in the last step an interchange of $\sum$ and $\int$ has been performed, and the 'translation identity' has been used (see the wiki page). Shift the integral and and you'll get $$\sum_{k=0}^n \binom{n}{k}\frac{B_k(x)}{n-k+1}= \int_{x}^{x+1}B_n(u) \,du = x^n$$ where another ID from the wiki page has been used (which can be taken as a definition). If this is for a homework problem, then you'd probably want to prove the identities in this proof.

ADDENDUM: Here's a proof of the 'translation theorem' using generating functions. (It's assumed the reader is familiar with the generating function for the Bernoulli polynomials.) $$\quad (T) \quad \sum_{k=0}^n \binom{n}{k}\, u^{n-k} \,B_k(x) = B_n(x+u) $$ It's easy to see that $$\quad (*) \quad \frac{t\,e^{t\,x}}{e^t-1} e^{t\,u} = \frac{t\,e^{t\,(x+u)}}{e^t-1}.$$ On the left-hand side (LHS), make a Cauchy product: $$ LHS(*)=\sum_{k=0}^\infty \frac{t^k}{k!}B_k(x)\cdot \sum_{m=0}^\infty \frac{t^m}{m!}u^m = \sum_{n=0}^\infty \frac{t^n}{n!} \sum_{k=0}^n \binom{n}{k}\, u^{n-k} \,B_k(x).$$ On the RHS, use the generating function again. Formula (T) follows by equating coefficients of $t.$