Sum over the side or Vertical of Pascal's triangle / Infinite Series of Binomial coefficients summed over UPPER index Is there any closed form formula of even an equivalent for this binomial infinite series :
$$
F_k(x) = \sum_{n=k}^{\infty} \binom{n}{k}   x^{n} 
$$
in which $|x|<1$ and k is a given integer ?
For example, $F_3(x) = x^3 + 4x^4 +10 x^5 +20 x^6 + 35 x^7 +... $
We all know that $ (1+x)^{n} = \sum_{k=1}^{n} \binom{n}{k} x^{k} $
But it's not the same summation : in the above k is fixed and n varies from k to infinity...
In Pascal's triangle, it's a sum aver a parallel to the side of the triangle, while the classical one goes along its base.
 A: Take any $k∈ℕ^+$ and $x∈ℂ$ such that $|x|<1$.
$F_k(x)·(1-x) = \sum_{n=k}^∞ \binom{n}{k}x^n - \sum_{n=k+1}^∞ \binom{n-1}{k}x^n$
$= \sum_{n=k}^∞ \binom{n-1}{k-1}x^n = x·\sum_{n=k-1}^∞ \binom{n}{k-1}x^n$
$= F_{k-1}(x)·x$.
Therefore $F_k(x) = x^k·(1-x)^{-k-1}$, since $F_0(x) = (1-x)^{-1}$. Done.
~ ~ ~
Note that it is inadvisable to use differentiation of power series unless necessary, because it is not rigorous unless you invoke some uniform convergence theorems (which are much more complicated than what I used here). It is also bad to use logarithms because of the issues with branch cuts.
Finally, this is closely related to using the forward difference operator for indefinite summation.
A: If you write the first few terms, you have:
$$S_k(x) = \frac{F_k(x)}{x^k} = \binom k k + \binom{k+1}kx + \binom{k+2}kx^2 + \cdots$$
Which becomes:
$$S_k(x) = 1 + (k+1)x + \frac{(k+1)(k+2)}{2!}x^2 + \cdots$$
So you're looking for the function that:
$$S_k(0) = 1$$
$$S'_k(0) = k+1$$
$$S''_k(0) = (k+1)(k+2)$$
$$S^{(n)}_k(0) = (k+1)(k+2)\cdots(k+n)$$
EDIT
As pointed out in the comments by Robearz, the function we're looking for is simply:
$$S_k(x) = (1 - x)^{-(k + 1)}$$
So:
$$F_k(x) = x^k(1 - x)^{-(k + 1)}$$
A: $$\begin{align*}{T(x)} &= \binom k k + \binom{k+1}kx + \binom{k+2}kx^2 + \cdots\\&=\binom k k + \frac{k+1}{1}\binom{k}kx + \frac{k+2}{2}\binom{k+1}kx^2 + \cdots
\\&=xT(x)+\binom k k + \frac{k}{1}\binom{k}kx + \frac{k}{2}\binom{k+1}kx^2 + \cdots\end{align*}$$
$$\implies (1-x)T(x)=\binom k k + \frac{k}{1}\binom{k}kx + \frac{k}{2}\binom{k+1}kx^2 + \cdots$$
Differentiating both sides
$$\implies (1-x)T'(x)-T(x)= kT(x)$$
$$\implies (1-x)T'(x)= (k+1)T(x)$$
$$\frac{dT(x)}{T(x)}=\frac{(k+1)dx}{1-x}$$
$$\implies \ln(T(x))=-(k+1)\ln(1-x)+C$$
$$\implies T(x)(1-x)^{k+1}=C$$
$T(0)=1$
$$\implies T(x)=\frac1{(1-x)^{k+1}}$$
$$\implies F_k(x)=x^kT(x)=\frac{x^k}{(1-x)^{k+1}}$$
