Asymptotic behaviour of some series I'm wondering how to examine the asymptotic behaviour as $x$ goes to $0^{+}$ for the following series (which is the continuous function in the right neighbourhood of $0$):
$$\sum_{n=1}^{\infty}\sin^{2}\left(\frac{n\pi}{2}\right)\exp\left(-\frac{n^2\pi^2 x}{4}\right)$$ 
i.e. how to find a simple continuous function $g$ such that the above series is equivalent to $g(x)$ as $x$ goes to $0^{+}$. Thank you for any replies!
 A: For every positive $u$, consider the two series 
$$
s(u)=\sum\limits_{n=1}^{+\infty}\mathrm e^{-u^2n^2},\qquad
r(u)=\sum\limits_{n=0}^{+\infty}\mathrm e^{-u^2(2n+1)^2}.
$$
Since the function $t\mapsto e^{-t^2}$ is decreasing on $t\geqslant0$, for every $n\geqslant1$,
$$
\int_{n}^{n+1}\mathrm e^{-u^2t^2}\mathrm dt\leqslant\mathrm e^{-u^2n^2}\leqslant\int_{n-1}^n\mathrm e^{-u^2t^2}\mathrm dt.
$$
Summing these, one gets
$$
\int_{1}^{+\infty}\mathrm e^{-u^2t^2}\mathrm dt\leqslant s(u)
\leqslant
\int_{0}^{+\infty}\mathrm e^{-u^2t^2}\mathrm dt
=
\frac1u\int_{0}^{+\infty}\mathrm e^{-t^2}\mathrm dt=\frac{\sqrt{\pi}}{2u}.
$$
The LHS is greater than the RHS minus $1$ hence, for every positive $u$,
$$
\frac{\sqrt{\pi}}{2u}-1\leqslant s(u)\leqslant \frac{\sqrt{\pi}}{2u},
$$
which is more than enough to see that $us(u)\to \frac12\sqrt{\pi}$ when $u\to0$.
The series $r(u)$ is liable to the same treatment, which yields $r(u)\sim \dfrac{\sqrt{\pi}}{4u}$. Using $u^2=\pi^2x/4$, this yields
$$
S(x)=\sum_{n=1}^{\infty}\sin^{2}\left(\frac{n\pi}{2}\right)\exp\left(-\frac{n^2\pi^2 x}{4}\right)=r(u)\sim\frac1{2\sqrt{\pi x}}.
$$
One sees that this method yields the stronger result that, for every positive $x$,
$$
\frac1{2\sqrt{\pi x}}-1\leqslant S(x)\leqslant\frac1{2\sqrt{\pi x}}.
$$
A: Not sure how to actually derive it, but I can get the answer with Mathematica at least. Note $f(x) = g(e^{-\pi^2 x})$ where $g(z) = \sum_{k=0}^\infty z ^ {(k+1/2)^2}$. The latter is a special case of an Elliptic Theta  function. Then since $f(x) \rightarrow \infty$ slower than $1/x$, I made an educated guess on the leading order, so evaluating with Mathematica, it turns out that $\lim_{x\rightarrow 0^+} f(x)x^{1/2}=\frac{1}{2 \sqrt{\pi}}$. (Links go to Wolfram Alpha). So $f(x) \sim \frac{1}{2 \sqrt{\pi x}}$ as $x \rightarrow 0^+$.
A: The sine factor only means that you only take odd sum indices.
The sum should be split into a finite initial part that is interpreted as a Riemann sum and scales to an approximation of the integral $\int_{0}^{big}c(\exp(-a t^2))dt$.
It will obviously take some work to choose the cut-off point correctly, so that both the rest of the sum and the rest of the integral to infinity are small.
Then you can use the known result for $\int_0^{\infty}\exp(-t^2)dt$ to conclude.
A: Your series can be expressed by means of the theta function
$$\theta(x)\ :=\ \sum_{n=-\infty}^\infty e^{-n^2\pi x}\ .$$
Indeed, one has
$$\eqalign{f(x):=\sum_{n\geq1,{\rm odd}}\ \exp\Bigl({-n^2\pi^2 x\over 4}\Bigr)&= 
\sum_{n\geq1}\ \exp\Bigl({-n^2\pi^2 x\over 4}\Bigr) -
\sum_{m\geq1}\ \exp\Bigl({-4m^2\pi^2 x\over 4}\Bigr) \cr &={1\over2}\Bigl(\theta\bigl({\pi x\over 4}\bigr) -\theta(\pi x)\Bigr)\ .\cr}$$
Now this theta function satisfies a famous functional equation (the "Jacobi transformation"), proven in Fourier analysis:
$$\theta(x)={1\over\sqrt{x}}\ \theta\Bigl({1\over x}\Bigr)\qquad(x>0)\ .$$
Therefore we get
$$f(x)={1\over 2\sqrt{\pi x}}\Biggl(2\theta\Bigl({4\over\pi x}\bigr)-\theta\Bigl({1\over \pi x}\Bigr)\Biggr)\ .$$
Taking $k=0$ in the theta series this not only confirms the result $g(x)={1\over 2\sqrt{\pi x}}$  of other answers. In addition, our formula shows that the approximation is unbelievably good when $x\to 0$: The terms corresponding to $k=\pm1$ in the two theta series are of order $e^{-4/x}$ resp. $e^{-1/x}$.
