Please explain how we get the equation $\sum_{n=0}^{\infty} n![(n+1)B_{n+1}-B_n]=-1$ Please explain how we get the equation $$\sum_{n=0}^{\infty} n![(n+1)B_{n+1}-B_n]=-1$$ from the equation $$\sum_{n=0}^{\infty} n![(n+1)x-1]x^n=-1, \ x \in \Bbb Z, \ \  ........(1)$$
Here $B_n$ are Bernouli numbers, $B_0=1, \ B_1=-1/2, \ B_3=0, B_4=1/6, \cdots.$
I have seen this within an article which  claims as follows:
At first put $x=1$ in $(1)$ to get 
$\sum_{n=0}^{\infty} n!n=-1$ 
and then  put $x=-1$ to get
$\sum_{n=0}^{\infty} n!(-1)^n(n+2)=1$.
Then the article claims $\sum_{n=0}^{\infty} n![(n+1)B_{n+1}-B_n]=-1$.
But how does the process goes in?
I could not understand the trick behind the claim starting from $(1)$.
Can you please check the claim? 
 A: Part $\text{(C)}$ is the solution.
$\text{(A)}$
A possibility to show the consistency of the equation. 
$\displaystyle f(x):=\sum\limits_{n=0}^\infty((n+1)!x^{n+1}-n!x^n)~$ with $~f(0)=-1$ 
$x(xf(x))'=\sum\limits_{n=1}^\infty((n+1)!x^{n+1}-n!x^n)=f(x)-(x-1)$
This works for $~f(x)=-1~$ under the condition $~f(0)=-1$ .  

$\text{(B)}~~$ For all $~x\in\mathbb{N} :$
$\displaystyle –x = \sum\limits_{k=0}^{x-1}f(k) = \sum\limits_{n=0}^\infty\left( (n+1)!\frac{B_{n+2}(x)-B_{n+2}(0)}{n+2} - n!\frac{B_{n+1}(x)-B_{n+1}(0)}{n+1} \right)$ 
If it's also correct for $~x\in\mathbb{R} :$ 
$\displaystyle -1 = \frac{d}{dx}\sum\limits_{n=0}^\infty\left((n+1)!\frac{B_{n+2}(x)-B_{n+2}(0)}{n+2} - n!\frac{B_{n+1}(x)-B_{n+1}(0)}{n+1}\right)$
$\displaystyle\hspace{0.7cm} =\sum\limits_{n=0}^\infty ((n+1)!B_{n+1}(x)-n!B_n(x))$
It remains to show, that $~x\in\mathbb{R}~$ can be used instead of $~x\in\mathbb{N}~$ .
Perhaps the reason lies in the fact that the sum $~\sum\limits_{k=0}^{x-1}f(k) ~$ is a polynomial 
(here: of degree one) that is clearly defined by enough but finally many 
(here: two) interpolation points.

$\text{(C)}~~$ Solution. 
We use analytic continuation: $~\displaystyle\sum\limits_{k=1}^\infty k^n =\zeta(-n) = -\frac{B_{n+1}}{n+1}~$ , $~n\in\mathbb{N}_0$
$\displaystyle -1 = (f(x)-x+1)' = \sum\limits_{n=1}^\infty((n+1)!(n+1)x^n-n!nx^{n-1}) $
Sum up from $~k=1~$ to $~\infty~$ :
Left side: $~\sum\limits_{k=1}^\infty (-1) = -\zeta(0) = B_1$ 
Right side: 
$\displaystyle\sum\limits_{k=1}^\infty (f(x)-x+1)'|_{x=k} = $
$\hspace{1cm}\displaystyle =\sum\limits_{k=1}^\infty \sum\limits_{n=1}^\infty((n+1)!(n+1)k^n-n!nk^{n-1})$
$\hspace{1cm}\displaystyle = \sum\limits_{n=1}^\infty \left((n+1)!(n+1)\left(\sum\limits_{k=1}^\infty k^n\right)-n!n\left(\sum\limits_{k=1}^\infty k^{n-1}\right)\right) $
$\hspace{1cm}\displaystyle = \sum\limits_{n=1}^\infty ((n+1)!(n+1)\zeta(-n)-n!n \zeta(1-n))$
$\hspace{1cm}\displaystyle = -\sum\limits_{n=1}^\infty ((n+1)!B_{n+1}-n!B_n)$
It follows:
$\displaystyle \sum\limits_{n=1}^\infty ((n+1)!B_{n+1}-n!B_n) = -B_1~~~ | +(B_1-B_0)$
$\displaystyle \sum\limits_{n=0}^\infty ((n+1)!B_{n+1}-n!B_n) = -B_0 = -1$
q.e.d. :)
