How to associate Bernouli numbers or Bernouli polynomials into the relation
\begin{equation} \sum_{n=0}^{\infty} \left[n(4x-1)+(2x) \right]x^n=0, \ \ \ \cdots
\cdots (1) \end{equation} and \begin{equation} \sum_{n=0}^{\infty} \left[n^2(4x-1)^2+(2x+12x^2) \right]x^n=0, \ \ \ \cdots
\cdots (2) \end{equation}
My approach:
The Bernouli numbers are
$B_0=0, \ B_1=-1/2, \ B_2=1/6, \ B_4=-1/30, \cdots$ and $B_{2n+1}=0,n =1,2,3,…..$,
where as the Bernouli polynomials are
$B_0(x)=1, \\
B_1(x)=x-1/2, \\
B_2(x)=x^2-x+1/6 \\
B_3(x)=x^3-3/2x^2+1/2x, \cdots$.
I think Bernouli polynomials can be used in $(1)$ and $(2)$ but I could not.
Please help me.
Next,
I have seen a note where it is given \begin{eqnarray} \sum_{n=0}^{\infty} n![(n+1)x-1]x^n=-1 \end{eqnarray} From this equation,putting $x=1, \ x=-1$
\begin{eqnarray} \sum_{n=0}^{\infty} n!nx^n=-1, \ x=1 \\
\sum_{n=0}^{\infty} n!(-1)^n(n+2)=1, \ x=-1 \end{eqnarray}
and then it writes \begin{eqnarray}\\
\sum_{n=0}^{\infty} n![(n+1)B_{n+1}-B_n]=1, \ ……..(*) \end{eqnarray}, where $B_n$ are Bernouli numbers in $(*)$.
I did not understand the process of associating Bernouli numbers.
Can I use the same process in $(1)$ and $(2)$ ?
Kindly help me
Best Answer
The result you may be looking for is $$ \sum_{n=0}^k n! ((n+1) B_{n+1} - B_n) = (k+1)! B_{k+1} - 1. \tag{1}$$ I don't see the left side converging as $\,k\to\infty.$ Similarly also we get $$ \sum_{n=0}^k n![(n+1)x-1]x^n = (k+1)!x^{k+1} -1. \tag{2}$$ These are both example of telescoping sums. The step from equation $(2)$ to equation $(1)$ is a classical one using the Umbral calculus with the linear mapping defined by $$ L: x^n \mapsto B_n. \tag{3} $$