Prove uniform tightness

Question

Given $X_{n}\sim N(\sin(n\pi),1)$, how could we prove $X_{n}=O_{p}(1)$?

Thanks!

Answer 1

$Y=X_n-\cos (n\pi) \sim N(0,1)$. So $P(|X_n| >M)\leq P(|Y|>M-|\cos (n \pi))\leq P(|Y| >M-1) <\epsilon$ if $M$ is sufficiently large. [BTW $\cos (n\pi) =(-1)^{n}$].