What is the sum of $\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$?

Question

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.

Answer 1

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.

Answer 2

If $|x|<1$ :

The series converges by the ratio test

$\displaystyle\sum_{n=1}^\infty (n^2+n^3)x^{n-1}=\frac{x+1}{(1-x)^3}+\frac{x^2+4x+1}{(1-x)^2(x^2-2x+1)}=\frac{4 x + 2}{x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1}$

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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

Answer 3

$$ \dfrac{d}{dx}x\dfrac{d^2}{dx^2}x^{n+1} = (n^2+n^3)x^{n-1} $$ maybe this will help?

Answer 4

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.

Answer 5

Another way to solve it....

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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.

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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 $n=0$.

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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).