Limit relevant to parametrised semi-group 2

Question

Let $s\geq 1, \epsilon >0, T>0$ and $f \in \mathcal{C}([0,T], H^s(\mathbb{T}))$. Define the function $$g(t,x):= \int_0^t(Id-\exp\left(-i\tau\epsilon \Delta)\right)f(\tau,x)d\tau.$$ I want to prove that $\lim_{\epsilon \rightarrow 0}\|g\|_{L^\infty(0,T; H^s)}= 0.$ To this end,write first

$$\|g(t,.)\|^2_{H^s} := \sum_{n\in \mathbb{Z}}(1+n^2)^s\left| \int_0^t (1-\exp(i\tau\epsilon n^2))\hat{f}(\tau,n)d\tau\right|^2.$$ "My idea" is to cut the sum into two terms:

1. The first one is $\sum_{n\leq N}(1+n^2)^s\left| \int_0^t (1-\exp(i\tau\epsilon n^2))\hat{f}(\tau,n)d\tau\right|^2.$ Here we have finite sum. Hence, it suffices to use $$\lim_{\epsilon \rightarrow 0}\left(\sup_{\tau}|1-\exp(i\tau\epsilon n^2)|\right) =0, n\leq N.$$
2. The second term is $a(t,N):=\sum_{n> N}(1+n^2)^s\left| \int_0^t (1-\exp(i\tau\epsilon n^2))\hat{f}(\tau,n)d\tau\right|^2.$ I need to prove that $$\lim_{N\rightarrow +\infty}\sup_{t\leq T}a(t,N)= 0.$$

Thank you for your hints.