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I want to express the following series as the value of an elementary function

$$\sum_{n=1}^{\infty}n^2x^n$$

I was doing some research, and it seems that to solve this kind of exercises I must consider $f$ as $$f(x)=\frac{g(x)}{(1-x)^p}$$ Then I simply have to find the function $g(x)$ and the value of $p$; however I don't know how to do this, any suggestions?

Átila Correia
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Wrloord
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    Also https://math.stackexchange.com/questions/338852/find-a-closed-form-of-the-series-sum-n-0-infty-n2xn?noredirect=1 or https://math.stackexchange.com/questions/2150495/finding-a-function-corresponding-to-the-complex-power-series?noredirect=1 or https://math.stackexchange.com/questions/562286/putting-power-series-in-closed-form?noredirect=1 – Gary Mar 25 '22 at 04:46

1 Answers1

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Let $f:(-1,1)\to\mathbb{R}$ be defined by $$f(x)=\frac{x}{1-x}=\frac1{1-x}-1.$$ We have that $$f(x)=\sum_{n=1}^{\infty}x^n,$$ and the convergence is uniform (I will not prove the uniform convergence here). Therefore, $$f'(x)=\sum_{n=1}^{\infty}\frac{\mathrm{d}}{\mathrm{d}x}x^n=\sum_{n=1}^{\infty}nx^{n-1},$$ so $$f_1(x)=xf'(x)=\sum_{n=1}^{\infty}nx^n.$$ This series also converges uniformly, and again, I will not prove that here. Thus $$f_1'(x)=\sum_{n=1}^{\infty}n^2x^{n-1},$$ and so $$f_2(x)=xf_1'(x)=\sum_{n=1}^{\infty}n^2x^n.$$ Notice that $$f'(x)=\frac1{(1-x)^2},$$ so $$f_1(x)=\frac{x}{(1-x)^2}=\frac{1-(1-x)}{(1-x)^2}=\frac1{(1-x)^2}-\frac1{1-x}.$$ Thus $$f_1'(x)=\frac2{(1-x)^3}-\frac1{(1-x)^2}$$ and $$f_2(x)=\frac{2x}{(1-x)^3}-\frac{x}{(1-x)^2}=\frac{2[1-(1-x)]}{(1-x)^3}-\frac{1-(1-x)}{(1-x)^2}=\frac2{(1-x)^3}-\frac3{(1-x)^2}+\frac1{1-x}.$$

In fact, this can be made more general. Consider the sequence of functions defined by the recurrence $$f_0(x)=\frac1{1-x}-1,$$ $$f_{n+1}(x)=xf_n'(x).$$ Then, for all $k\in\mathbb{N},$ $$f_k(x)=\sum_{n=1}^{\infty}n^kx^n.$$ You will find that $$f_k(x)=P_k[g(x)],$$ where $$g(x)=\frac1{1-x}$$ and $P_k$ is a polynomial function of degree $k+1.$ Here, $$P_0(x)=x-1,$$ and $$f_{k+1}(x)=P_{k+1}[g(x)]=x(P_k\circ{g})'(x)=xP_k'[g(x)]g'(x)=xg(x)^2P_k'[g(x)]=[1-(1-x)]g(x)^2P_k'[g(x)]=\left[1-\frac1{g(x)}\right]g(x)^2P_k'[g(x)]=[g(x)^2-g(x)]P_k'[g(x)],$$ so $$P_{k+1}(x)=(x^2-x)P_k'(x)=xP_0(x)P_k'(x)=[P_0(x)^2-P_0(x)]P_k'(x).$$

You will also find that these functions are related to the polylogarithms. Take a look at https://en.wikipedia.org/wiki/Polylogarithm. In fact, you will see that $$f_k(x)=\mathrm{Li}_{-k}(x).$$ Also, these functions have an important relationship to the $\Theta$ operator. The $\Theta$ operator is defined as $$\Theta[f]=id\cdot{D}[f],$$ where $D$ is the derivative operator, and $id$ is the identity function, where for all $x$ in some subset of the real numbers, $$id(x)=x.$$ The sequence I defined above is such that $$\Theta[f_n]=f_{n+1},$$ such that the operator acts as a sort of shifting operator or ladder operator for this particular sequence of functions. This also gives an explicit expression for $f_n$ in terms of $f_0,$ as $$f_n=\Theta^n[f_0].$$

Angel
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