Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a method for evaluating infinite series of the form:

$\displaystyle\sum_{k=1}^{\infty}\frac{1}{a^{k}+1}, \;\ a\in \mathbb{N}$?. For instance, say $a=2$ and we have


I know this converges to $\approx .7645$, but is it possible to find its sum using some clever method?.

It would seem the Psi function is involved. I used the sum on Wolfram and it gave me

$$-1+\frac{{\psi}_{1/2}^{(0)}\left(1-\frac{\pi\cdot i}{\ln(2)}\right)}{\ln(2)}$$

I am familiar with the Psi function, but I am unfamiliar with that notation for Psi. What does the 1/2 subscript represent?

share|cite|improve this question
The integer a should be assumed to be at least 2 (and not simply in N). – Did Aug 28 '11 at 13:07
@Cody: I suggest that you don't call this function the "psi function" as there are many many functions for which we use the symbol $\psi$. Instead, call it by its name, the q-Polygamma function. (This makes it easier to look up too) Or in this case, the q-digamma function. – Eric Naslund Aug 28 '11 at 14:10
Not a proof but I've tried "Plot[NSum[1/(a^k + 1), {k, 1, [Infinity]}] - (-Log[a/(a - 1)] + QPolyGamma[0, 1 - (I [Pi])/Log[a], 1/a])/Log[a], {a, 2, 10}]" in Mathematica and it turns out to be a zero function. – ziyuang Aug 28 '11 at 15:00
$\frac12$ in your expression for the q-polygamma function would be the value of the "quantum parameter" q. – J. M. Aug 28 '11 at 16:48
up vote 4 down vote accepted

What you want to look at is Lambert Series. Notice that in the expression Wolfram Alpha spat out, the $\frac{-\pi i}{\log 2}$ is simply there to turn the positive sign into a negative one so that we are dealing with Lambert Series.

Theses series are power series where the coefficients are given by Dirichlet convolution, so they are often related to multiplicative functions.

Let $$L(q)=\sum_{n=1}^\infty a_n \frac{q^n}{1-q^n}=\sum_{n=1}^\infty \left(\sum_{d|n} a_d \right) q^n.$$

Then we can rewrite a series related to yours above: $$\sum_{n=1}^\infty \frac{1}{2^n-1}=\sum_{n=1}^\infty \frac{d(n)}{2^n}.$$

However, if the coefficients $a_n$ are functions which become nice when convolved with $1$ we can get something different. For example $$\sum_{n=1}^\infty \mu(n)\frac{q^n}{1-q^n}=q.$$

share|cite|improve this answer

Not entirely relevant, but it's known that these numbers are irrational.
Peter B. Borwein, On the irrationality of $\sum_{n\gt0}(1/(q^n+r))$, J. Number Theory 37 (1991), no. 3, 253–259, MR1096442 (92b:11046) proves the sum is irrational when $r$ is rational and $q\ge2$ is an integer. Could be worth a look.

EDIT: As Dan notes, $r$ can't be zero. Also, it can't be of the form $-q^m$, lest we find ourselves dividing by zero.

share|cite|improve this answer
r cannot be 0 if the sum is irrational. – Dan Brumleve Aug 29 '11 at 5:41

This is not precisely the question asked, but still very interesting: A sum evaluated in terms of Jacobi theta functions... $$ \sum_{n={-\infty}}^\infty \frac{x^n}{1-a q^{2n}} = \frac{\theta_1(i\ln(ax)/2,q)\theta_2(0,q)\theta_3(0,q)\theta_4(0,q))}{2i\theta_1(i\ln(x)/2,q)\theta_1(i\ln(a)/2,q)} $$ for $|q|^2 < |x| < 1$.

Also: $$ \sum_{n=-\infty}^\infty \frac{1}{\alpha^n-a\beta^n} = \frac{\theta_1(i\ln(\alpha/a)/2,\sqrt{\beta/\alpha}) \theta_2(0,\sqrt{\beta/\alpha})\theta_3(0,\sqrt{\beta/\alpha}) \theta_4(0,\sqrt{\beta/\alpha})}{2i \theta_1(i\ln(\alpha)/2,\sqrt{\beta/\alpha}) \theta_1(i\ln(a)/2,\sqrt{\beta/\alpha})} $$ for $0 < |\beta| < 1 < |\alpha|$.

(Provided I copied it correctly .)

share|cite|improve this answer
The two formulas are the same, using $x=1/\alpha$ and $q^2=\beta/\alpha$. – Did Aug 28 '11 at 14:43
Well, theta functions and $q$-functions are rather intimately related, so I suppose this is not so surprising... – J. M. Aug 28 '11 at 16:49
Where did you copy it from? – anon Aug 28 '11 at 19:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.