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You’ve chosen the right method, it should be conclusive.
$e = \sum_{k=0}^{+\infty} \frac{1}{k!} $
$\implies 2\pi n! e = 2\pi*K + \frac{2\pi}{n+1} + o(\frac{1}{n})$, K being a positive integer (why is that?)
$\sin(2\pi n! e)= sin(\frac{2\pi}{n+1} + o(\frac{1}{n})) = \frac{2\pi}{n+1} + o(\frac{1}{n}) $
$\implies a_n = \frac{2n}{n+1} + o(1) $
Hence (if I’m not mistaken): $(a_n) \rightarrow 2$ , when $ n \rightarrow +\infty$
Hint: $$\frac1{n+1}+n!\sum_{k=0}^n\frac1{k!}\lt n!\mathrm e\lt\frac1n+n!\sum_{k=0}^n\frac1{k!}$$