$f(t)=\sum_{n=1}^{\infty}2\sin^{3}( \frac{\pi n}{2})\exp(\frac {-n^{2}\pi^{2}t}{2}).$ find $a,b $ such that $f(t)\sim ae^{-bt}$

Question

Consider a function $$f(t)=\sum_{n=1}^{\infty}2\sin^{3}(\pi n/2)\exp(-n^{2}\pi^{2}t/2).$$ I want to find two constants $a,b $ such that $f(t)\sim ae^{-bt}$ as $t\rightarrow\infty.$

I observed that the $f(t)$ is actually alternating: $$f(t)=2\Big(\exp(-\pi^{2}t/2)-\exp(-9\pi^{2}t/2)+\cdots\Big),$$ so my guess is that as $t$ goes to $\infty$, all the terms except the first one will decay faster so goes to $0$, so my choice should be $a=2$ and $b=-\pi^{2}/2$, but I don't know how to show it formally...

Thanks for any help!

Answer 1

Let $u:=e^{-\pi^2t/2}$. Your series is

$$u-u^9+u^{25}-\cdots(-1)^ku^{(2k+1)^2}+\cdots$$

For $t\to\infty$, we have $u\to0$ and the series can be reduced to its first term.

For instance, when $u=0.5$, the second term is only $0.39\%$ of the first and as the series is alternating, the error by omitting all these terms is smaller.

Caution anyway, when $u$ nears $1$ ($t$ small), the function has a very bad behavior.

Below the six first partial sums.