Approximating an infinite series for large values of $h$ I'm working through a textbook on the statistical physics of polymers and came across this partition sum for a polymer chain confined between two walls:
$$\chi(h)=\frac{8}{\pi^2}\sum_{i=1,3,5..}\frac{1}{i^2}e^{-i^2\frac{\pi^2 R_\text{g}^2}{h^2}}$$
The textbook states that for large $h$ this function can be approximated by
$$\chi(h)=1-\frac{4R_\text{g}}{ h\sqrt{\pi}}$$
I've tried a few methods and cannot seem to reproduce this result, would someone be so kind to help me on the way.
 A: Let's consider $$\chi(h)=\frac{8}{\pi^2}\sum_{i=1,3,5..}\frac{1}{i^2}e^{-i^2\frac{\pi^2 R_\text{g}^2}{h^2}}=\chi(\alpha)=\frac{8}{\pi^2}\sum_{i=1,3,5..}\frac{1}{i^2}e^{-\alpha i^2}$$ where $\alpha=\frac{\pi^2 R_\text{g}^2}{h^2}\ll1$
We denote $$S_0(\alpha)=-\frac{\pi^2}{8}\frac{\partial}{\partial\alpha}\chi(\alpha)=\sum_{i=1,3,5..}e^{-\alpha i^2}=\sum_{i=1,2,3..}e^{-\alpha i^2}-\sum_{i=2,4..}e^{-\alpha i^2}=\sum_{i=1,2,3..}\big(e^{-\alpha i^2}-e^{-4\alpha i^2}\big)$$
Let's also introduce
$\,\,\theta(x)=\sum_{n=-\infty}^\infty e^{-\pi xn^2}$
Then
$$S_0(\alpha)=\frac{1}{2}\Big(\theta\big(\frac{\alpha}{\pi}\big)-1\Big)-\frac{1}{2}\Big(\theta\big(\frac{4\alpha}{\pi}\big)-1\Big)$$
We can use the functional equation for theta-function, which states (http://www.math.columbia.edu/~woit/fourier-analysis/theta-zeta.pdf , page 4):
$$\theta(x)=\sum_{n=-\infty}^\infty e^{-\pi xn^2}=\frac{1}{\sqrt x}\theta \big(\frac{1}{x}\big)=\frac{1}{\sqrt x}\sum_{k=-\infty}^\infty e^{-\frac{\pi k^2}{x}}$$
Therefore, we can write
$$S_0(\alpha)=\frac{1}{2}\Big(\sqrt\frac{\pi}{\alpha}\theta\big(\frac{\pi}{\alpha}\big)-1\Big)-\frac{1}{2}\Big(\sqrt\frac{\pi}{4\alpha}\theta\big(\frac{\pi}{4\alpha}\big)-1\Big)$$
$$=\frac{\sqrt\pi}{2\sqrt\alpha}\big(1+2e^{-\frac{4\pi^2}{\alpha}}+...\big)-\frac{\sqrt\pi}{2\sqrt{4\alpha}}\big(1+2e^{-\frac{4\pi^2}{4\alpha}}+...\big)$$
Dropping the exponentially small terms,
$$S_0(\alpha)\sim\frac{\sqrt\pi}{4\sqrt\alpha}$$
$$\chi(\alpha)=-\frac{8}{\pi^2}\int^\alpha S_0(t)dt +C\sim C-\frac{8}{\pi^2}\frac{\sqrt\pi}{2}\sqrt\alpha$$
The constant of integration is defined as
$$C=\chi(0)=\frac{8}{\pi^2}\sum_{i=1,3,5..}\frac{1}{i^2}=1$$
Therefore,
$$\chi(h)\sim1-\frac{8}{\pi^2}\frac{\sqrt\pi}{2}\sqrt\alpha=1-\frac{4}{\sqrt\pi}\frac{R_g}{h}$$
In fact, this is a very good approximation, because other terms of the asymptotics are exponentially small $\big(\sim \frac{h}{R_g}e^{-\frac{h}{R_g}}\big)$
