We have,
$I_n=\int \limits_0^{\pi / 2} x^n \cos x d x$
$\Rightarrow \quad I_n =\left[x^n \sin x\right]_0^{\pi / 2}-\int \limits_0^{\pi / 2} n x^{n-1} \sin x d x$
$\Rightarrow \quad I_n =\left(\frac{\pi}{2}\right)^n-\left[n x^{n-1}(-\cos x)\right]_0^{\pi / 2}$
$\Rightarrow \quad I_n =\left(\frac{\pi}{2}\right)^n-n(n-1) I_{n-2}^{\pi / 2} n(n-1) x^{n-2} \cos x d x$
$\Rightarrow \quad I_n+n n(n-1) I_{n-2}=\left(\frac{\pi}{2}\right)^n$
$\text { Now, } \sum \limits_{n=2}^{\infty}\left[\frac{I_n}{n !}+\frac{I_{n-2}}{(n-2) !}\right]$
$\Rightarrow \quad \sum \limits_{n=2}^{\infty}\left[\frac{I_n}{n !} n(n-1) I_{n-2}\right]$
$\sum \limits_{n=2}^{\infty} \frac{\left(\frac{\pi}{2}\right)^n}{n !}=e^{\pi / 2}-1-\frac{\pi}{2}$
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