By integration, differentiation, or any other valid operation : I am asked to find the function which these represent
$$\sum_1^\infty n^2x^n$$
I tried doing everything - but I just can't do it. What I've done so far :
The series is convergent $<=>$ $-1<x<1$ And uniformly convergent on a compact set $K \subset (-1,1)$ So I can use the property of uniformly convergent series to use differentiation integration etc. But I couldn't move a pencil. Any help will be appreciated
On
We may consider that $$ f(x,t)\stackrel{\text{def}}{=}\sum_{n\geq 1} e^{nt} x^n = \frac{x e^t}{1-x e^t} \tag{1}$$ as soon as $\left|x e^t\right|<1$. By applying $\frac{\partial^2}{\partial t^2}$ to both sides, then evaluating at $t=0$, $$ \sum_{n\geq 1} n^2 x^n = \left.\frac{\partial^2}{\partial t^2}\left(\frac{x e^t}{1-x e^t}\right)\right|_{t=0} = \left.\frac{x e^t(1+x e^t)}{(1-x e^t)^3}\right|_{t=0}=\color{red}{\frac{x(1+x)}{(1-x)^3}}.\tag{2}$$ On the other hand, since $n^2$ is a second-degree polynomial in the $n$ variable, we know in advance that $$ (1-x)^3 \sum_{n\geq 1} n^2 x^n = \sum_{n\geq 1}\left(p(n)-3p(n-1)+3p(n-2)+p(n-3)\right) x^n$$ is a polynomial in $x$ with degre $\leq 2$.
On
Let $x=e^{-\theta}$, then $$\sum_{n\in\mathbb{N}}n^{2}x^n=\sum_{n\in\mathbb{N}}n^{2}e^{-\theta{n}}=$$ $$=\frac{\partial^{2}}{\partial{\theta}^{2}}\sum_{n\in\mathbb{N}}e^{-\theta{n}}=\frac{\partial^{2}}{\partial{\theta}^{2}}[\frac{1}{1-e^{-\theta}}-1]=$$ $$=-\frac{\partial}{\partial{\theta}}\frac{e^{-\theta}}{(1-e^{-\theta})^{2}}=$$ $$=\frac{e^{-\theta}(1-e^{-\theta})^{2}+2(1-e^{-\theta})e^{-2\theta}}{(1-e^{-\theta})^{4}}=$$ $$=\frac{x}{(1-x)^{2}}+\frac{2x^{2}}{(1-x)^{3}}$$
Note that $$ x\frac{d}{dx} x^n = nx^n, $$ so this can be found by applying $xd/dx$ twice to $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x} $ in the region of absolute convergence.