I need to determine an asymptotic expansion when $q \rightarrow 1$ of the sum

$$S(q)=\sum_{n=0}^{\infty} \frac{q^n}{ (q^n + 1)^2 }.$$

Numerical computations suggest that $S(q)\sim\frac{c}{|q-1|}$ with $c \approx 0.5$.

  • $\begingroup$ "for n→∞ and q→1" Simultaneously? How? $\endgroup$ – Did Nov 16 '13 at 18:01
  • $\begingroup$ Thank you for your comment. Actually I need first to take the limit $n \rightarrow \infty$ and then $q \rightarrow 1$. $\endgroup$ – Vahagn Poghosyan Nov 16 '13 at 18:14
  • $\begingroup$ @Did Thank you for the editing of my question. I would like to note that the function $S(q)$ is defined also for $q>1$. $\endgroup$ – Vahagn Poghosyan Nov 16 '13 at 18:34

It is indeed true that $S(q) \sim \frac{1/2}{|1-q|}$. We'll use the same method as in this answer.

First assume $0 < q < 1$. The terms of the sum are strictly decreasing in $n$, so

$$ \int_0^\infty \frac{q^x}{(q^x + 1)^2}\,dx \leq \sum_{n=0}^{\infty} \frac{q^n}{(q^n+1)^2} \leq \frac{1}{4} + \int_0^\infty \frac{q^x}{(q^x + 1)^2}\,dx. $$

To evaluate the integral, make the change of variables $q^x = y$ to get

$$ \begin{align} \int_0^\infty \frac{q^x}{(q^x + 1)^2}\,dx &= -\frac{1}{\log q} \int_0^1 \frac{dy}{(y+1)^2} \\ &= -\frac{1}{2\log q}. \end{align} $$


$$ \sum_{n=0}^{\infty} \frac{q^n}{(q^n+1)^2} \sim -\frac{1}{2\log q} $$

as $q \to 1^-$. Of course $\log q \sim q-1$ as $q \to 1$, so this is equivalent to

$$ \sum_{n=0}^{\infty} \frac{q^n}{(q^n+1)^2} \sim \frac{1}{2(1-q)} $$

as $q \to 1^-$.

To address the case when $q > 1$ we note that $S(q) = S(1/q)$, so that

$$ S(q) \sim \frac{1}{2(1-\frac{1}{q})} = \frac{q}{2(q-1)} \sim \frac{1}{2(q-1)} $$

as $q \to 1^+$. Thus

$$ S(q) \sim \frac{1}{2|1-q|} $$ as $q \to 1$.


Suppose that $q>1$ and re-write the sum as $$S(q) = \frac{1}{4} + \sum_{n\ge 1} \frac{q^n}{(q^n+1)^2}$$ The sum term is harmonic and may be re-written as $$T(x) = \sum_{n\ge 1} \frac{e^{nx}}{(e^{nx}+1)^2}$$ which we will evaluate at $x=\log q.$

We will evaluate $T(x)$ by inverting its Mellin transform. Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad g(x) = \frac{e^x}{(e^x+1)^2}.$$ We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \frac{e^x}{(e^x+1)^2} x^{s-1} dx = \int_0^\infty \frac{e^{-x}}{(1+e^{-x})^2} x^{s-1} dx = \int_0^\infty \sum_{m\ge 1} (-1)^{m+1} m e^{-mx} x^{s-1} dx\\ \sum_{m\ge 1} (-1)^{m+1} m \int_0^\infty e^{-mx} x^{s-1} dx = \Gamma(s) \sum_{m\ge 1} \frac{(-1)^{m+1} m}{m^s} = \Gamma(s) \sum_{m\ge 1} \frac{(-1)^{m+1}}{m^{s-1}}\\ = \Gamma(s) \left(1-\frac{1}{2^{s-2}}\right) \zeta(s-1).$$ Therefore the Mellin transform $Q(s)$ of $T(x)$ is given by $$Q(s) = \Gamma(s) \left(1-\frac{1}{2^{s-2}}\right) \zeta(s-1) \zeta(s).$$ The Mellin inversion integral for $Q(s)$ is $$\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds.$$ The two zeta function terms taken together cancel the poles of the gamma function, so we are left with just two poles/residues: $$\mathrm{Res}(Q(s)/x^s; s=1) = \frac{1}{2x} \quad\text{and}\quad \mathrm{Res}(Q(s)/x^s; s=0) = -\frac{1}{8}.$$ Therefore in a neighborhood of zero we have that $$T(x) \sim \frac{1}{2x}-\frac{1}{8}.$$ Now as $q$ goes to one $\log q$ goes to zero, so this expansion at $x=\log q$ definitely applies. We get that $$S(q) = \frac{1}{4} + \frac{1}{2x}-\frac{1}{8} = \frac{1}{8} + \frac{1}{2\log q}.$$ This approximation is better than the conjecture from the original question. Now use that for $q$ close to one, $$\frac{1}{2\log q} \sim \frac{1}{2(q-1)} + \frac{1}{4} - \frac{1}{24} (q-1)+\cdots$$ to conclude that $$S(q) \sim \frac{3}{8} + \frac{1}{2(q-1)}.$$

  • $\begingroup$ Thank you for the answer. Note that we can obtain the constant term and also the next terms, if we use Euler–Maclaurin formula. This formula is a generalization of the Integral test for convergence used by @Antonio Vargas. $\endgroup$ – Vahagn Poghosyan Nov 17 '13 at 11:14
  • 1
    $\begingroup$ This is true. If you want a reference for this observation consult page 50 of the technical report "Mellin Transform Asymptotics" by Flajolet and Sedgewick. $\endgroup$ – Marko Riedel Nov 17 '13 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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