Please explain how we get the equation $\sum_{n=0}^{\infty} n![(n+1)B_{n+1}-B_n]=-1$

Question

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?

Answer 1

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 $~f(0)=-1$ .

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$\text{(B)}~~$ 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.

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$\text{(C)}~~$ 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. :)