<s>[INST] What is the sum of $\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$? Consider the power sequence
$$\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$$
What is the function to which it sums to?
My reasoning is to differntiate the sum with respect to $x$, then to integrate with respect to x from $0$ to $x$ after variation of the sum into a Taylor series form.
I mean:
$$\int _0 \frac{d}{dx}\sum_{n=1}^\infty (n^2+n^3)x^{n-1}dx=\int_0 \sum_{n=1}^\infty \frac{n^2+n^3}{n-1}x^{n-2}dx$$
But it doesn't seem to work in this exercise.
 [/INST] If $|x|<1$ :
The series converges by the ratio test
$\displaystyle\sum_{n=1}^\infty (n^2+n^3)x^{n-1}=\frac{x+1}{(1-x)^3}+\frac{x^2+4x+1}{(1-x)^2(x^2-2x+1)}=\frac{4 x + 2}{x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1}$

If $|x|>1$ :
The series diverges by the ratio test
$\displaystyle\sum_{n=1}^\infty (n^2+n^3)x^{n-1}=\sum_{n=1}^\infty\frac{n^2x^n}x+\sum_{n=1}^\infty\frac{n^3x^n}x=\sum_{n=1}^{\infty} n^{2} x^{n - 1} \left(n + 1\right)$ is the best I can come with
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