Infinite factorial summation without the factorial

It'd be better if gave a proof of that. The proof is simple though.

$\sum_{k=0}^\infty\binom{k+2}{k}x^k=\sum_{k=0}^\infty\binom{k+2}{2}x^k=\sum_{k=0}^\infty\frac{1}{2!}(k+2)(k+1)x^k$

Differentiating twice the identity stated below w.r.t $x$ gives us the closed form for the sum above.

$\sum_{k=0}^\infty x^{k+2}=\frac{1}{1-x}-x-1\quad\forall~x:0\lt |x|\lt 1$

---

The problem posed is the sum stated at the top of this comment with $x=\left(-\frac 13\right)$ which satisfies the convergence condition as $0\lt \left|-\frac 13\right|=\frac 13\lt 1$ .