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I have

$$ \sum_{k=1}^{\infty} k^3 x^k $$

I know, if we start with

$$ \frac{ 1 }{1-x } = \sum_{k \geq 0} x^k $$

and differentiate this three times and multiply by $x$ when appropriate, we can find a closed form for the given power series. But, it seems like very laborious computation of rational functions.

Qs: Is there a clever way to find a closed form for this power series?

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Let $S(x)=\sum_{k=1}^\infty x^k=\frac{x}{1-x}$. Then, we have

$$\begin{align} S'(x)&=\sum_{k=1}^\infty kx^{k-1}\\\\ xS'(x)&=\sum_{k=1}^\infty kx^{k}\\\\ (xS'(x))'&=\sum_{k=1}k^2x^{k-1}\\\\ x(xS'(x))'&=\sum_{k=1}k^2x^{k}\\\\ (x(xS'(x))')'&=\sum_{k=1}k^3x^{k-1}\\\\ x(x(xS'(x))')'&=\sum_{k=1}k^3x^{k} \end{align}$$

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Considering $$\sum_{k=1}^{\infty} k^3 x^k$$ I think that an easy way is to write $$k^3=k(k-1)(k-2)+a k(k-1)+b k=k^3+(a-3) k^2+ (b-a+2)k$$ from which $a=3$ and $b=1$. So $$\sum_{k=1}^{\infty} k^3 x^k=\sum_{k=1}^{\infty}k(k-1)(k-2)x^k+3\sum_{k=1}^{\infty}k(k-1)x^k +\sum_{k=1}^{\infty}kx^k $$ $$\sum_{k=1}^{\infty} k^3 x^k=x^3\sum_{k=1}^{\infty}k(k-1)(k-2)x^{k-3}+3x^2\sum_{k=1}^{\infty}k(k-1)x^{k-2} +x\sum_{k=1}^{\infty}kx^{k-1} $$ $$\sum_{k=1}^{\infty} k^3 x^k=x^3 \left(\sum_{k=1}^{\infty} x^k\right)'''+3x^2 \left(\sum_{k=1}^{\infty} x^k\right)''+x \left(\sum_{k=1}^{\infty} x^k\right)'$$

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