Evaluating a class of infinite sums in closed form

Posted on by John

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(z) = \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-positive 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.

Update: Here is a proof that the sum is an integer when n = 2. From a comment by Theophylline on Substack.

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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3 thoughts on “Evaluating a class of infinite sums in closed form”

  3. (sorry about the telegraphic nature of the last comment. I messed up the paste). Meant to say: Good stuff. For more fun check out A = B, by Marko Petkovsek, Herbert S Wilf, Doron Zeilberger, A K Peters/CRC Press, 1996.

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