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.
 A: $$
\dfrac{d}{dx}x\dfrac{d^2}{dx^2}x^{n+1} = (n^2+n^3)x^{n-1}
$$
maybe this will help?
A: Here's a method which uses finite differences and generating functions. Let $a_{n}=(n+2)(n+1)^2$, so that the power series $$\sum_{n=0}^\infty a_n x^n=\sum_{n=1}^\infty a_{n-1} x^{n-1}=\sum_{n=1}^\infty (n+1)n^2 x^{n-1}$$ may be recognized as being the same as that of the OP. Let's focus our attention on the integer sequence  $\{a_n\}$. Since the terms are polynomials in $n$, we compute the first few orders of finite differences:
\begin{array}{lllcccccc}
&(\Delta_0 a)_n=a_n &:\quad &2, &12, &36, &80,  &\ldots&  n^3+n^2,&\cdots\\
&(\Delta_1 a)_n=(\Delta_0 a)_n-(\Delta_0 a)_{n-1} &:\,&2, &10, &24, &44,  &\ldots&  3n^2-n,&\cdots\\
&(\Delta_2 a)_n=(\Delta_1 a)_n-(\Delta_1 a)_{n-1} &:\,&2, &8, &14, &20, &\ldots&  6n-4,&\ldots\\
&(\Delta_3 a)_n=(\Delta_2 a)_n-(\Delta_2 a)_{n-1} &:\,&2, &6, &6, &6, &\ldots&  6,&\cdots\\
&(\Delta_4 a)_n=(\Delta_3 a)_n-(\Delta_3 a)_{n-1} &:\,&2, &4, &0, &0, &\ldots&  0,&\cdots\\
\end{array}
Thus the fourth differences all vanish except for the first two terms, consistent with $a_n$ being cubic in $n$. We can turn this around: If we start with the sequence $\{2,4,0,0,\ldots\}$ and repeatedly take cumulative sums, we obtain the other sequences going from bottom to top. In particular, $a_n$ is the 'fourth cumulative sum' of the initial sequence.
Why does this matter? First, note that the generating function of the sequence $\{2,4,0\ldots\}$ is simply $2+4x$. Moreover, there is a simple way to implement 'cumulative sums' with generating function: Just divide by $(1-x).$ (Check it for yourself if you're not sure.) So the generating function of the fourth cumulative sum is just $\boxed{\sum\limits_{n=0}^\infty a_n x^n =\dfrac{2+4x}{(1-x)^4}}$, and we conclude that this is the desired series summation.
A: This is a very general approach.
$$n^3+n^2 = n(n-1)(n-2) + 4n(n-1) + 2n$$
So $$(n^3+n^2)x^n = \left(x^3\left(\frac{d}{dx}\right)^3 + 4x^2\left(\frac{d}{dx}\right)^2 + 2x\frac{d}{dx}\right)x^n$$
So $$\sum (n^3+n^2)x^n =\left(x^3\left(\frac{d}{dx}\right)^3 + 4x^2\left(\frac{d}{dx}\right)^2 + 2x\frac{d}{dx}\right)\frac{1}{1-x}$$
In general, if $(n)_k = n(n-1)\cdots(n-k+1)$ is the falling factorial, then these are a basis for all polynomials, so if $p(n)$ is a polynomial of degree $d$, then we can write:
$$p(n)=\sum_{k=0}^{d} a_k(n)_k$$
Then $$\sum_{n=0}^\infty p(n)x^n = \left(\sum_{k=0}^d a_kx^k\left(\frac{d}{dx}\right)^k\right)\frac{1}{1-z}$$
Now $$\left(\frac{d}{dx}\right)^k\frac{1}{1-x} = \frac{k!}{(1-x)^{k+1}}$$
So that gives:
$$\sum_{n=0}^\infty p(n)x^n = \sum_{k=0}^d \frac{k!a_kx^k}{(1-x)^{k+1}}=\frac{\sum_{k=0}^d k!a_kx^k(1-x)^{d-k}}{(1-x)^{d+1}}$$
This shows that $(1-x)^{d+1}\sum_{n=0}^\infty p(n)x^n$ is a polynomial of degree at most $d$. The beauty of this is that you can just multiply polynomials:
$$(1-x)^{d+1}\sum_{n=0}^d p(n)x^n$$ and ignore the terms of degree bigger than $d$ to figure out what the numerator polynomial is.
So if $p(n)=n^3+n^2$ then 
$$\begin{align}(1-x)^4(2x+12x^2+36x^3) &= 2x(1-4x+6x^2-4x^3+x^4)(1+6x+18x^2) \\
&= 2x(1+2x +0x^2 +\dots)
\end{align}$$
And your series is:
$$\sum_{n=1}^\infty (n^3+n^2)x^{n-1} = \frac{1}{x}\sum_{n=0}^\infty p(n)x^n=\frac{2(1+2x)}{(1-x)^4}$$
Interesting to note that $(1-x)^4\sum_{n=0}^d p(n)x^d$ is essentially the same as the approach in Semiclassical's answer - multiplying by $(1-x)$ repeatedly is essentially the same as doing his finite differences.
A: Another way to solve it.... 

First: 
Note that 
$$\frac{d^2}{dx^2}\Bigl(nx^{n+1}\Bigr)=n^2(n+1)x^{n-1}=(n^2+n^3)x^{n-1}\qquad(1)$$
we will use this later.

Second: 
We know that
$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\qquad|x|<1$$
Taking the derivative respect to $x$, we have:
$$\begin{array}{rcl}
\frac{d}{dx}\Bigg(\sum_{n=0}^{\infty}x^n\Bigg)&=&\frac{d}{dx}\Bigg(\frac{1}{1-x}\Bigg)\\
\sum_{n=0}^{\infty}\frac{d}{dx}x^n&=&\frac{1}{(1-x)^2}\\
\sum_{n=1}^{\infty}nx^{n-1}&=&\frac{1}{(1-x)^2}\\
\sum_{n=0}^{\infty}nx^{n-1}&=&\frac{1}{(1-x)^2}\qquad(2)
\end{array}$$
where in the LHS of $(2)$ I summed a $0$, that is equivalent to the term for $n=0$.

Third:
Take $(2)$ and multiply by $x^2$:
$$\sum_{n=0}^{\infty}nx^{n+1}=\frac{x^2}{(1-x)^2}\qquad|x|<1\qquad(3)$$
Now we have the general term $nx^{n+1}$ in the sum, so if we derive respect to $x$ two times, we will have our function because of the relation of $(1)$. Then:
$$\begin{array}{rcl}
\frac{d^2}{dx^2}\Bigg(\sum_{n=0}^{\infty}nx^{n+1}\Bigg)&=&\frac{d^2}{dx^2}\Bigg(\frac{x^2}{(1-x)^2}\Bigg)\\
\sum_{n=1}^{\infty}\frac{d^2}{dx^2}\Big(nx^{n+1}\Big)&=&\frac{d}{dx}\Bigg(\frac{-2x}{(1-x)^3}\Bigg)\\
\sum_{n=1}^{\infty}(n^2+n^3)x^{n-1}&=&\frac{2(2x+1)}{(1-x)^4}\\
\end{array}$$
Note we had to drop the constant term that happens at $n=0$ for the 1st derivative of the series (second line).
A: 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
