$\lim_{\epsilon\to0}\frac{\cos(\epsilon-n\frac{\pi}{2})}{\epsilon^n}$

Question

We were doing generalized integrals in class and this integral came out. I tried using integration by parts and got something repeating. We're gonna let $\epsilon \rightarrow 0$ and $x\rightarrow\infty$
$$\int_0^{\infty}\frac{-\cos t}{t}dt=\Big[\frac{-\sin t}{t}\Big]_{\epsilon}^{x}+\Big[\frac{\cos t}{t^2}\Big]_{\epsilon}^{x}+\Big[\frac{2\sin t}{t^3}\Big]_{\epsilon}^{x}+\Big[\frac{-6\cos t}{t^4}\Big]_{\epsilon}^{x}+\int_{\epsilon}^x\frac{-24\cos t}{t^5}dt$$
$$=\sum_{n=1}^{\infty}\Big[\frac{(-1)^{n+1}(n-1)!\cos{(t-n\frac{\pi}{2})}}{t^n}\Big]^x_{\epsilon}$$
$$=\sum_{n=1}^{\infty}(-1)^{n+1}(n-1)!\Big(\lim_{x\to\infty}\frac{\cos(x-n\frac{\pi}{2})}{x^n}-\lim_{\epsilon\to0}\frac{\cos(\epsilon-n\frac{\pi}{2})}{\epsilon^n}\Big)$$
$$=\sum_{n=1}^{\infty}(-1)^{n}(n-1)!\lim_{\epsilon\to0}\frac{\cos(\epsilon-n\frac{\pi}{2})}{\epsilon^n}$$
The first limit is $0$ by squeeze theorem but what about the second one, I don't know how to start.
Please share your work, thank you for your time!

Answer 1

Hint: $$\int_0^{\arccos{(0.5)}}\frac{\cos{(t)}}{t} dt\gt \int_0^{\arccos{(0.5)}}\frac{0.5}t dt\to\infty$$