The other day I ran across the surprising identity

sum_{n=1}^infty frac{n^3}{2^n} = 26

and wondered how many sums of this form can be evaluated in closed form like this. Quite a few it turns out.

Sums of the form

sum_{n=1}^infty frac{n^k}{c^n}

evaluate to a rational number when k is a non-negative integer and c is a rational number with |c| > 1. Furthermore, there is an algorithm for finding the value of the sum.

The sums can be evaluated using the polylogarithm function Lis(z) defined as

text{Li}_s(x) = sum_{n=1}^infty frac{z^n}{n^s}

using the identity

sum_{n=1}^infty frac{n^k}{c^n} = text{Li}{-k}left(frac{1}{c}right)

We then need to have a way to evaluate Lis(z). This cannot be done in closed form in general, but it can be done when s is a negative integer as above. To evaluate Li−k(z) we need to know two things. First,

Li_1(z) = -log(1-z)

and second,

text{Li}_{s-1}(z) = z frac{d}{dz} text{Li}_s(z)

Now Li0(z) is a rational function of z, namely z/(1 − z). The derivative of a rational function is a rational function, and multiplying a rational function of z by z produces another rational function, so Lis(z) is a rational function of z whenever s is a non-negative integer.

Assuming the results cited above, we can prove the identity

sum_{n=1}^infty frac{n^3}{2^n} = 26

stated at the top of the post.The sum equals Li−3(1/2), and

text{Li}_{-3}(z) = left(z frac{d}{dz}right)^3 frac{z}{1-z} = frac{z(1 + 4z + z^2)}{(1-z)^4}

The result comes from plugging in z= 1/2 and getting out 26.

When k and c are positive integers, the sum

sum_{n=1}^infty frac{n^k}{c^n}

is not necessarily an integer, as it is when k = 3 and c = 2, but it is always rational. It looks like the sum is an integer if c= 2; I verified that the sum is an integer for c = 2 and k = 1 through 10 using the PolyLog function in Mathematica. The sum is occasionally an integer for larger values of c. For example,

sum_{n=1}^infty frac{n^4}{3^n} = 15

and

sum_{n=1}^infty frac{n^8}{3^n} = 17295

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