For computing the present worth of an infinite sequence of equally spaced payments $(n^{2})$ I had the need to evaluate

$$\displaystyle\sum_{n=1}^{\infty}\frac{n^{2}}{x^{n}}=\dfrac{x(x+1)}{(x-1)^{3}}\qquad x>1.$$

The method I used was based on the geometric series $\displaystyle\sum_{n=1}^{\infty}x^{n}=\dfrac{x}{1-x}$ differentiating each side followed by a multiplication by $x$, differentiating a second time and multiplying again by $x$. There is at least a second (more difficult) method that is to compute the series partial sums and letting $n$ go to infinity.

Question: Is there a closed form for

$$\displaystyle\sum_{n=1}^{\infty }\dfrac{n^{p}}{x^{n}}\qquad x>1,p\in\mathbb{Z}^{+}\quad ?$$

What is the sketch of its proof in case it exists?


The general closed form is

$$\displaystyle \sum_{k=1}^{\infty} k^n x^k = \frac{1}{(1 - x)^{n+1}} \left( \sum_{m=0}^{n} A(n, m) x^{m+1} \right)$$

where $A(n, m)$ are the Eulerian numbers. When I have time I will edit with a few more details. If you only want the answer for a particular small value of $n$ then see Section 3 of my notes on generating functions. I will also mention that for a particular value of $n$ one can deduce the answer by using the identity

$$\displaystyle \sum_{k=0}^{\infty} {k+n \choose n} x^k = \frac{1}{(1 - x)^{n+1}}$$

and writing $k^n$ as a linear combination of the polynomials ${k+r \choose r}$ (in $k$), for example using a finite difference table.

  • $\begingroup$ Changing $x\longmapsto 1/x$ and for the exponent $n=2$, this formula gives the same as above! $\sum_{k=1}^{\infty }k^{2}x^{-k}=\frac{1}{( 1-x^{-1})^{3}}( A(2,0)x^{-1}+A(2,1)x^{-2}+A(2,2)x^{-3}) $ $=\frac{1}{\left( x-1\right) ^{3}}( x^{2}A(2,0)+xA(2,1)+A(2,2)) $ $=\frac{1}{\left( x-1\right) ^{3}}( x^{2}+x+0) $ $\endgroup$ – Américo Tavares Sep 9 '10 at 0:26
  • $\begingroup$ For $n=3$ the sum is $\displaystyle\sum_{k=1}^{\infty}k^{3}x^{k}=\dfrac{x^{3}+4x^{2}+x}{(1-x)^{4}}$ $\endgroup$ – Américo Tavares Sep 9 '10 at 14:12
  • $\begingroup$ @Américo Tavares: to clarify, the fact that the coefficients A(n, m) exist can be deduced from general principles, if that is what you mean by "closed form": just repeatedly differentiate with respect to x and multiply by x and you can prove this by induction. What is surprising is that the coefficients A(n, m) are non-negative integers with nice combinatorial interpretations. $\endgroup$ – Qiaochu Yuan Sep 9 '10 at 15:25
  • $\begingroup$ @Qiaochu Yuan: I am not sure if "closed form" is the correct designation for what I was thinking about. I just wanted to know whether, and how, the series might be evaluated in terms of elementary (rational) functions. Your reply assures it. I used the repeated differentiation and multiplication method to evaluate the $n=3$ case. $\endgroup$ – Américo Tavares Sep 9 '10 at 15:53
  • 1
    $\begingroup$ One nice fact is that the sequence of coefficients in the numerator is always palindromic. $\endgroup$ – Robin Chapman Sep 9 '10 at 20:34

Here is a different look:

The differentiating and multiplying by $x$ gives rise to Stirling Numbers of the Second Kind.

Say you denote the operator of differentiating and multiplying by $x$ as $D_{x}$

Then we have that

$$(D_{x})^{n}f(x) = \sum_{k=1}^{n} s(n,k) f^{(k)}(x) x^{k}$$

where $s(n,k)$ is the stirling number of the second kind and $f^{(k)}(x)$ is the $k^{th}$ derivative of $f(x)$.

This can easily be proven using the identity $$s(n,k) = s(n-1,k-1) + k \cdot s(n-1,k)$$

Here is a table for the Stirling numbers of the second kind (from the wiki page):

n/k   0     1     2     3     4      5      6      7      8      9
0     1
1     0     1
2     0     1     1
3     0     1     3     1
4     0     1     7     6     1
5     0     1     15    25    10     1
6     0     1     31    90    65     15     1
7     0     1     63    301   350    140    21     1
8     0     1     127   966   1701   1050   266    28     1
9     0     1     255   3025  7770   6951   2646   462    36     1

So in your case, we can start with $ f(x) = \frac{1}{1-x}$ and obtain that

$$ \sum_{k=0}^{\infty} k^{n} x^k = \sum_{r=1}^{n} r! \cdot s(n,r) \frac{x^{r}}{(1-x)^{r+1}}$$

For example, in your case for $n=3$ we get

$$\sum_{k=0}^{\infty} k^3 x^k = \frac{1! \cdot 1 \cdot x}{(1-x)^2} + \frac{2! \cdot 3 \cdot x^2}{(1-x)^3} + \frac{3! \cdot 1 \cdot x^3}{(1-x)^4}$$

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

$$ = \frac{x^3 + 4x^2 + x}{(1-x)^4} $$

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    $\begingroup$ Nice approach. I'm a bit partial to using $\left\{\begin{matrix}n\\\\k\end{matrix}\right\}$ instead of $s(n,k)$, but still, +1. $\endgroup$ – J. M. is a poor mathematician Sep 9 '10 at 21:40

According to mathworld The $n^{th}$ moment of the geometric distribution with parameter $p$, which is $ \sum p (1-p)^n n^k $ can be expressed in terms of the polylogarithm function.

  • 1
    $\begingroup$ To be precise, the polylogarithm is given by the infinite series $\mathrm{Li}_n(z)=\sum_{j=1}^\infty \frac{z^j}{j^n}$ ; you can transform the series for the nth moment to match the form of the polylogarithm. $\endgroup$ – J. M. is a poor mathematician Sep 9 '10 at 0:31

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