Concept about $\binom{ \infty}k$ concept about function $\binom{ \infty}k$
For 
$ k \in \mathbb{N}$

The idea of ​​this function is derived from my power sum formula

link for my power sum formula

Formula is as

$$\sum_{k=1}^{n} k^{m}=\sum_{b=1}^{m+1} \binom{n}b\sum_{i=0}^{b-1} (-1)^{i}(b-i)^{m}\binom{b-1}i$$

This formula helps to derive $\binom{ \infty}k$ function and to calculate it's value.  we know the negative values of zeta function.if

$$ \zeta(-m)=\lim_{n\to \infty}\sum_{k=1}^{n} k^{m}$$

So can we construct it as
 $$\zeta(-m)=\sum_{b=1}^{m+1} \binom{\infty}b\sum_{i=0}^{b-1} (-1)^{i}(b-i)^{m}\binom{b-1}i$$
Then we can calculate, if we substitute value $\zeta (0)=-1/2$ then$\binom{\infty}1=-1/2$

Again we can calculate next value using or substituting previous values of $\binom{\infty}k$.

Other values of $\binom{\infty}k$are

$$\binom{\infty}2=5/12$$
$$\binom{\infty}3=-3/8$$
$$\binom{\infty}4=251/720$$
$$...$$
And so on.

Application

Definition

Let's us define a sequence as :

$$a=(a_{1},a_{2},a_{3},...)$$

Difference between two term is as follows

$$\triangle^{0}a_{n}=a_{n}$$
$$\triangle^{1}a_{n}=a_{n+1}-a_{n}$$
More generally
$$\triangle^{m}a_{n}=\triangle^{m-1}a_{n+1}-\triangle^{m-1}a_{n}$$
If there exist some $m$ for $\triangle^{m}a_{n}=0$ such that $\forall n \in \mathbb{N}$

Then $$\sum_{k=1}^{n} a_{k}=\sum_{b=1}^{m+1} \binom{n}b\triangle^{b-1}a_1$$

now if $\lim_{n \to \infty}$ put values $\binom{\infty}b$

$$\sum_{k=1}^{\infty} a_{k}=\sum_{b=1}^{m+1} \binom{\infty}b\triangle^{b-1}a_1$$

And Get the result

Example

To calculate 
$1+3+5+...+(2n-1)+...=1/3$

Question


  
    Q1- how this function $\binom{\infty}k$ impact to understanding  and  analysis of mathematics?
    
    Q2- can we derive it's definition/algorithm to calculate  function $\binom{\infty}k$ for $k\in\mathbb{C}$  values by analysis in some field , i mean what is generalization for function $\binom{\infty}k$?
    
    Q3-Is it have some interesting properties?
  


Thank you very much for your suggestions comments and answer.
There is a unique polynomial $P_m(x)$ such that $P_m(0) = 0$ and $P_m(x)-P_m(x-1)= x^m$ so $P_m(n)=\sum_{k=1}^n k^m$ and you are setting $$P_m(\infty) = \zeta(-m)$$
For any polynomial $f(x)$ such that $f(0) = 0$ then $f(x)-f(x-1)$ is a polynomial $$f(x)-f(x-1) = \sum_{j=0}^d c_j x^j$$

You are setting $$f(\infty) = \sum_{j=0}^d c_j \zeta(-j)$$
With $g_l(x) = {x \choose l}$ then $$g_l(x)-g_l(x-1) =\sum_{j=0}^{l-1} b_{l,j} x^j$$
$${\infty \choose l} =\sum_{j=0}^{l-1} b_{l,j} \zeta(-j)$$
From $\sum_{n=1}^N (-1)^{n+1} = \frac{1+(-1)^{N+1}}{2}$ we have that $\eta(s)=(1-2^{1-s})\zeta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ is entire and $$\eta(-j) = \lim_{x \to 1} \sum_{n=1}^\infty (-1)^{n+1} n^j x^n$$
Let $$T (\sum_{j=0}^d c_j x^j)(x) = \sum_{j=0}^d \frac{c_j}{1-2^{1+j}} x^j$$
you are setting $$f(\infty) = \lim_{x \to 1} \sum_{n=1}^\infty (-1)^{n+1}x^n T(f(x)-f(x-1))(n) $$
The problem is that operator $T$ which makes it unobvious how to generalize your summation method to any sequences, not only polynomials.
There is a unique polynomial $P_m(x)$ such that $P_m(0) = 0$ and $P_m(x)-P_m(x-1)= x^m$ so $P_m(n)=\sum_{k=1}^n k^m$ and you are setting $$P_m(\infty) = \zeta(-m)$$
For any polynomial $f(x)$ such that $f(0) = 0$ then $f(x)-f(x-1)$ is a polynomial $$f(x)-f(x-1) = \sum_{j=0}^d c_j x^j$$

You are setting $$f(\infty) = \sum_{j=0}^d c_j \zeta(-j)$$
With $g_l(x) = {x \choose l}$ then $$g_l(x)-g_l(x-1) =\sum_{j=0}^{l-1} b_{l,j} x^j$$
$${\infty \choose l} =\sum_{j=0}^{l-1} b_{l,j} \zeta(-j)$$
From $\sum_{n=1}^N (-1)^{n+1} = \frac{1+(-1)^{N+1}}{2}$ we have that $\eta(s)=(1-2^{1-s})\zeta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ is entire and $$\eta(-j) = \lim_{x \to 1} \sum_{n=1}^\infty (-1)^{n+1} n^j x^n$$
Let $$T (\sum_{j=0}^d c_j x^j)(x) = \sum_{j=0}^d \frac{c_j}{1-2^{1+j}} x^j$$
you are setting $$f(\infty) = \lim_{x \to 1} \sum_{n=1}^\infty (-1)^{n+1}x^n T(f(x)-f(x-1))(n) $$
The problem is that operator $T$ which makes it unobvious how to generalize your summation method to any sequences, not only polynomials.