I'm trying to prove the following identity:
$$\sum_{k=0}^n \dfrac {\binom n k B_k(x)} {(n-k+1)} = x^n$$
I transformed this identity as follow:
$$\dfrac{1}{(n+1)}\sum_{k=0}^n \binom {n+1} k B_k(x) = x^n$$
Also I tried to do the following:
$$\sum_{k=0}^n \dfrac {C_n^k B_k(x)}{(n-k+1)} = n!\sum_{k=0}^n \dfrac {B_k(x)}{k!} \dfrac {1}{(n+1-k)!}$$
I add and subtract the (n+1)th summand:
$$n!\sum_{k=0}^{n+1} \dfrac {B_k(x)}{k!} \dfrac {1}{(n+1-k)!}-\dfrac{B_{n+1}(x)}{n+1}$$
Denote: $$ a_k= \dfrac {B_k(x)}{k!}, b_{n+1-k}=1$$ Hence:
$$A(t)=\sum_{k=0}^{\infty} \dfrac {B_k(x)}{k!}t^k=\dfrac {te^{xt}}{e^t-1}$$
$$B(t)=\sum_{k=0}^{\infty} \dfrac {t^k}{k!}=e^t$$
$$n!A(t)B(t)=n!\dfrac {e^{t(x+1)}t}{e^t-1}$$
I want to prove it: $$\sum_{k=0}^n \dfrac {\binom n k B_k(x)} {(n-k+1)} = x^n$$
So, I find the exponential generating function for the right side of the equality:
$$\sum_{k=0}^{\infty} \dfrac {(x^nt)^k}{k!}=e^{x^nt}$$
So I reformulated this task as follow.
Can we prove that:
$$n!\dfrac {e^{t(x+1)}t}{e^t-1}=\sum_{k=0}^{\infty} \dfrac {B_k(x)}{k}t^k +e^{x^nt}$$ Unfortunately, I can't find $$\sum_{k=0}^{\infty} \dfrac {B_k(x)}{k}t^k$$
Hope you can help me to prove this identity. Thanks for your attention!
Best Answer
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 $t.$