Find function represented by a power series

Question

I am looking for the function represented by the power series $\sum_{n=1}^\infty n^2z^n$

I am under the assumption that is converges to some radius R, and I can use the Cauchy-Hadamard Theorem to prove it but I am not sure how to find the function

Answer 1

Almost as dxiv commented, write $$A=\sum_{n=1}^\infty n^2 z^n=\sum_{n=1}^\infty (n(n-1)+n) z^n=z^2\sum_{n=1}^\infty n(n-1) z^{n-2}+z\sum_{n=1}^\infty n z^{n-1} $$ that is to say $$A=z^2 \left(\sum_{n=1}^\infty z^{n} \right)''+z \left(\sum_{n=1}^\infty z^{n} \right)'$$

Answer 2

We can also strictly follow @dxiv. Using the differential operator $D_z$ we apply the operator $zD_z$ to a series and get \begin{align*} \left(zD_z\right)\left(\sum_{n=0}^\infty a_nz^n\right)=z\sum_{n=0}^\infty na_nz^{n-1}=\sum_{n=0}^\infty na_n z^n\tag{1} \end{align*}

We obtain according to (1) by applying the operator $zD_z$ twice: \begin{align*} \color{blue}{\sum_{n=0}^\infty n^2 z^n}&=\left(zD_z\right)^2\left(\sum_{n=0}^\infty z^n\right)=\left(zD_z\right)^2\left(\frac{1}{1-z}\right)\\ &=\left(zD_z\right)\left(\frac{z}{(1-z)^2}\right)\\ &\,\,\color{blue}{=\frac{1+z}{(1-z)^3}} \end{align*}