# 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.

• the sum is over $n$ and not $x$ – hjpotter92 Sep 12 '14 at 14:20
• I see this. Maybe I didn't explain my method well. – E Be Sep 12 '14 at 14:21

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.

• +1 I noticed this method would work for a lot of series... Is there a particular name for this method? – CODE Sep 12 '14 at 15:54
• @CODE: Well, I synthesized it from what I know of generating functions and finite differences (mostly because I wanted to give a combinatoric meaning to the final result.). Likely it's got a specific name in that field. – Semiclassical Sep 12 '14 at 16:13
• This is very close to my answer, from a different approac. Generally, though $\Delta(a_0,a_1,\dots,a_n,\dots) = (a_1-a_0,a_2-a_1,\dots,a_{n+1}-a_n,\dots)$. Your $\Delta_1$ is essentially multiplying the series by $(1-x)$, keeping the $x^0$ term... – Thomas Andrews Sep 12 '14 at 16:40
• @Semiclassical. Nice, indeed. Thanks – Claude Leibovici Sep 12 '14 at 17:16
• @ThomasAndrews: Right, just approached from finite differences rather than derivatives. I actually thought about rewriting the finite differences in terms of just $(1-x)$ times the series, but decided to leave it in the more 'raw computation' form except for the last step to be more accessible. – Semiclassical Sep 12 '14 at 17:31

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

• Can u explain as to how – Shobhit Jan 25 '17 at 15:45
• @Shobhit Solving the original post and not the updated version.. $$\sum_{n=0}^\infty (n^2+n^3)x^{n-1} = \sum_{n=0}^\infty \dfrac{d}{dx}x\dfrac{d^2}{dx^2}x^{n+1}$$ Can permute the summation inside $$\dfrac{d}{dx}x\dfrac{d^2}{dx^2}\sum_{n=0}^\infty x^{n+1}$$ the latter summation is a geometric sum. Do you need me to continue? – Chinny84 Jan 25 '17 at 16:52
• Oh, got it. Thank u :) – Shobhit Jan 25 '17 at 18:39

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}

$$\sum_{n=1}^\infty (n^3+n^2)x^{n-1} = \frac{1}{x}\sum_{n=0}^\infty p(n)x^n=\frac{2(1+2x)}{(1-x)^4}$$

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.

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).

• $(d/dx)\frac 1{1-x}\neq \frac{x}{(1-x)^2}$. You added the $x4 back two steps early. – Thomas Andrews Sep 12 '14 at 15:48 • Ouch, thanks for noticing. Edited now. – cjferes Sep 12 '14 at 19:39 • You missed the point$\sum nx^n = x\sum nx^{n-1} = \frac{x}{(1-x)^2}$. It's just that the early steps, you re doing$\sum nx^{n-1}$. There should be an$x$, it just shouldn't be added when you did. – Thomas Andrews Sep 12 '14 at 19:44 • If I understand correctly what you're saying (that there's an error when derivating$(2)$), I am doing the whole thing again to avoid more errors, and I'll edit wherever is needed. Thanks again – cjferes Sep 12 '14 at 19:53 • (2) is actually wrong. There should be an$x$in the numerator, but not in the previous steps, which was my original complaint. So there should be an$x^2$numerator in the (3). And some of the derivatives on the right side strike me as suspect. You get the right answer, but (2) and (3) are wrong, so some of the rest of the argument must be wrong, too. – Thomas Andrews Sep 12 '14 at 19:55 If$|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