Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

After reading answers (especially M. Piau's) to my other question Asymptotic behaviour of some series, the more general question came to my mind, namely:
how to examine the asymptotic behaviour as $x\to 0^+$ for the following series:
$$\sum_{n=1}^\infty\sin^2\left(\frac{n\pi}{a}\right)\exp\left(-\frac{n^2\pi^2 x}{4}\right),$$ where $a>0$ is arbitrary, i.e. how to find a simple continuous function $g$ such that the above series is equivalent to $g(x)$ as $x$ goes to $0^+$.
I suspect that the answer would be again related to $\mathrm{const}\dfrac1{\sqrt x}$ but I just couldn't show it by using the approach of M. Piau. The problem now is that $\sin^2\left(\dfrac{n\pi}{a}\right)$ could take even infinitely many different values between $0$ and $1$. Once again thank you for any replies.

share|cite|improve this question

Let $S_a(x)$ denote the sum of this series and $S(x)$ the sum of the series without the sines. Obviously $S_a(x)\leqslant S(x)$. To get a lower bound, assume for a moment that $a\geqslant3/2$ and consider the set $N_a$ of positive integers $n$ such that $\sin^2(n\pi/a)\geqslant1/4$. Looking at the trigonometric circle and considering that the sequence of general term $n\pi/a$ makes jumps of size at most $2\pi/3$, which is the length of each of the intervals where $\sin^2\geqslant1/4$, one sees that the $k$th element of $N_a$ is at most $ka$ hence $$ 4S_a(x)\geqslant\sum_{n\in N_a}\exp\left(-n^2\pi^2 x/4\right)\geqslant\sum_{k=1}^{\infty}\exp\left(-k^2a^2\pi^2 x/4\right)=S(a^2x). $$ Since $S(x)\sim1/\sqrt{\pi x}$ when $x\to0$, $S_a(x)$ is asymptotically between two multiples of $1/\sqrt{x}$ when $x\to0$.

The argument may be adapted to any $a>1$. On the other hand, $S_1(x)=0$.

share|cite|improve this answer
Thank you very much for fast and as always very clear answer. – John Nov 14 '11 at 9:34
Tout le plaisir est pour moi. – Did Jan 13 '12 at 19:37

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.