Show that $\sum\limits_{n=0}^\infty \dbinom{n+2}{2}x^n$ converges for $|x|<1$ and calculate its sum.
My attempt:
$$\sum\limits_{n=0}^\infty \dbinom{n+2}{2}x^n=\sum\limits_{n=0}^\infty \frac{(n+2)(n+1)}{2!}x^n=\frac{1}{2}\sum\limits_{n=0}^\infty (n^2+3n+2)x^n$$
We look for absolute convergence of the power series:
$$\lim\limits_{n\longrightarrow\infty}\left|\frac{((n+1)^2+3(n+1)+2)x^{n+1}}{(n^2+3n+2)x^n}\right|<1$$
$$|x|\lim\limits_{n\longrightarrow\infty}\left|\frac{n^2+5n+6}{n^2+3n+2}\right|<1$$
$$|x|*1<1\Longleftrightarrow|x|<1$$
The power series converges $\forall x \in \mathbb{C}:|x|<1$
$\Box$
Now we calculate its sum:
$\frac{1}{2}\sum\limits_{n=0}^\infty (n+1)(n+2)x^n$
First of all, we know (for $|x|<1$):
$$\sum\limits_{k=0}^\infty x^k=\frac{1}{1-x}$$
Since the power series converges in a radius of $\rho=1$ we can differenciate elementwise:
$$\sum\limits_{k=0}^\infty \frac{d}{dx}x^k=\sum\limits_{k=0}^\infty kx^{k-1}=\frac{1}{(1-x)^2}$$
Differenciating another time gives us:
$$\sum\limits_{k=0}^\infty \frac{d}{dx}kx^{k-1}=\sum\limits_{k=0}^\infty k(k-1)x^{k-2}=\frac{2}{(1-x)^3}=\sum\limits_{k=-2}^\infty (k+2)(k+1)x^{k}$$
This gives us:
$$\frac{1}{2}\sum\limits_{n=0}^\infty (n+1)(n+2)x^n=\frac{1}{2}\left(\frac{2}{(1-x)^3}\right)=\frac{1}{(1-x)^3}=\sum\limits_{n=0}^\infty \dbinom{n+2}{2}x^n$$
It would be great if someone could look over it and check if my work is correct :) thank you