Bounding $\sum_{n=n_1}^\infty x^n (n+1)^2$

Question

I need to upperbound the sum $$\sum_{n=n_1}^\infty x^n (n+1)^2$$

where $0<x<1$ is a parameter. I know it can be done starting from
$$\sum_{n=n_1}^\infty x^n (n+1)^2\le \sum_{n=0}^\infty x^n (n+1)(n+2)-\sum_{n=0}^{n_1} x^n (n+1)(n+2)$$ using nasty calculations (integrating twice, then taking the second derivative from the closed form formula of the integral). I was wondering if this can be done in a simpler way.

Answer 1

Write $$ \sum_{n \in S} x^n(n+1)^2 = \sum_{n \in S} \frac{d}{dx} x^{n+1} (n+1) = \frac{d}{dx} \left( x \sum_{n \in S} x^n (n+1) \right) = \frac{d}{dx} \left( x \frac{d}{dx} \sum_{n \in S} x^{n+1} \right) $$ for any index set $S$. In the case that $S = \{n_1,n_1+1,\ldots\}$, this becomes \begin{align} \sum_{n = n_1}^\infty x^n(n+1)^2 &= \frac{d}{dx} \left( x \frac{d}{dx} x^{n_1 + 1}\sum_{n =0} x^n \right) \\&= \frac{d}{dx} \left( x \frac{d}{dx} \frac{x^{n_1 + 1}}{1-x} \right) \\&= \dfrac{{x}^{n_1}\left( x+{\left( 1+n_1 -n_1x \right) }^{2}\right) }{{\left( 1-x\right) }^{3}} \end{align}

Answer 2

Let $f(x)=\sum_m^\infty x^{n+2}$ and $g(x)=\sum_m^\infty x^{n+1}$. For $|x|\lt 1$ we have $$f''(x) -g'(x)=\sum_m^\infty (n+1)^2 x^n.$$

Note that $f(x)= \frac{x^{n+2}}{1-x}$ and $g(x)=\frac{x^{n+1}}{1-x}$.

Calculate $f''(x)-g'(x)$.