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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.

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1 Answer 1

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, $S_1(x)=0$.

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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

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