I have this function expressed as a series:
$$f(t) = \frac{1}{2} -\sum\limits_{n=1}^{\infty}\frac{2}{n\pi}\exp(-n^2\pi^2t)\sin\left(\frac{n\pi}{2}\right)$$
It comes from the Fourier series expansion of a function of two variables $(t, x)$ but I wanted to see the dependence for a single $x$ and small $t$.
I was about to proceed by saying "Since $t\ll1$ it follows that $\exp(-n^2\pi^2t)\approx1-n^2\pi^2t$", because for sufficiently small $t$ the higher order terms should vanish.
The problem is of course that $n$ is not bounded and consequently my approach doesn't work because no matter how small $t$ is, there will always be an $n$ that makes my linear approximation of the exponential wrong.
So, is there a way to correctly express the small $t$ approximation to $f(t)$ or do I necessarily have to take all the terms?
I thought that maybe the sign alternation in the $\sin(n\pi/2)$ can help some approximation be convergent, in the way normal convergence criteria are easier to evaluate with alternating signs.
Note that I'm an aspiring physicist, so I apologize for any mistakes made within the context of rigorous mathematical analysis. Thank you very much.
Poisson summation formula provides a way to handle small $t$ well. Let $y>0$ and $$S(x,y)=x+\sum_{n=1}^{\infty}\frac{2}{n\pi}e^{-(n\pi y)^2}\sin n\pi x,$$ so that $f(t)=1-S(1/2,\sqrt{t})$. Then the formula tells $S(x,y)=\sum\limits_{n=-\infty}^{\infty}s(x,y,n)$, where $$s(x,y,\omega)=\int_{-\infty}^{\infty}e^{-(\pi ty)^2-2\pi i\omega t}\frac{\sin\pi tx}{\pi t}\,dt$$ can be evaluated using "$\partial/\partial x$ trick" and known $\int_{-\infty}^{\infty}e^{-t^2-zt}\,dt=e^{z^2/4}\sqrt\pi$ for $z\in\mathbb{C}$ (!): $$s_x(x,y,\omega)=\frac{1}{2y\sqrt\pi}\left(e^{-\left(\frac{\omega}{y}+\frac{x}{2y}\right)^2}+e^{-\left(\frac{\omega}{y}-\frac{x}{2y}\right)^2}\right),$$ so that integration w.r.t. $x$, with noting that $s(0,y,\omega)=0$ clearly, gives $$s(x,y,\omega)=\frac12\operatorname{erf}\left(\frac{\omega}{y}+\frac{x}{2y}\right)-\frac12\operatorname{erf}\left(\frac{\omega}{y}-\frac{x}{2y}\right).$$ Thus, if $0<x<1$, then $\lim\limits_{y\to 0^+}S(x,y)=1$, and the "error" is $\mathcal{O}(e^{-(x/2y)^2}y/x)$, very small.