We have,
$$
I_n=\int_{\pi / 2}^{\infty} e^{-x} \cos ^n x d x
$$

Integration by parts
$$
\begin{gathered}
I_n=\left[-e^{-x} \cos ^n x\right]_{\pi / 2}^{\infty} \\
-\int_{\pi / 2}^{\infty}-e^{-x} \cdot n \cos ^{n-1} x(-\sin x) d x \\
=0-\int_{\pi / 2}^{\infty} n e^{-x} \cos ^{n-1} x \sin x d x
\end{gathered}
$$

By integration parts again, we get
$$
\begin{gathered}
=-n\left[\left(-e^{-x} \cos ^{n-1} x \sin x\right]_{\pi / 2}^{\infty}\right. \\
\left.-\int_{\pi / 2}^{\infty}-e^{-x}\left(\cos ^n x-(n-1) \cos ^{n-2} x \sin ^2 x\right) d x\right] \\
=-n\left[0+\int_{\pi / 2}^{\infty}\left(e^{-x} \cos ^n x-(n-1) e^{-x}\right.\right. \\
\left.\cos ^{n-2} x\left(1-\cos ^2 x\right) d x\right] \\
{\left[\because \sin ^2 x=1-\cos ^2 x\right]}
\end{gathered}
$$



$$
\begin{array}{cc}
=-n\left[\int_{\pi / 2}^{\infty} e^{-x} \cos ^n x d x-(n-1)\right. \\
& \left.\quad \int_{\pi / 2}^{\infty} e^{-x} \cos ^{n-2} x d x+(n-1) \int_{\pi / 2}^{\infty} e^{-x} \cos ^n x d x\right] \\
& =-n\left[I_n-(n-1) I_{n-2}+(n-1) I_n\right] \\
\Rightarrow & I_n=-n\left[n I_n-(n-1) I_{n-2}\right] \\
\Rightarrow & I_n=-n^2 I_n+n(n-1) I_{n-2} \\
\Rightarrow & I_n+n^2 I_n=n(n-1) I_{n-2} \\
\Rightarrow & I_n\left(n^2+1\right)=n(n-1) I_{n-2} \\
& \quad \frac{I_n}{I_{n-2}}=\frac{n(n-1)}{n^2+1}
\end{array}
$$

Put $\quad n=2018$
$$
\frac{I_{2018}}{I_{2016}}=\frac{2018(2018-1)}{(2018)^2+1}=\frac{2018 \times 2017}{(2018)^2+1}
$$