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
    Commented Nov 16, 2013 at 18:01
  • $\begingroup$ Thank you for your comment. Actually I need first to take the limit $n \rightarrow \infty$ and then $q \rightarrow 1$. $\endgroup$ Commented Nov 16, 2013 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 $q>1$. $\endgroup$ Commented Nov 16, 2013 at 18:34

2 Answers 2


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. $\endgroup$ Commented Nov 17, 2013 at 11:14
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    $\begingroup$ 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. $\endgroup$ Commented Nov 17, 2013 at 19:03

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