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Is it correct to squeeze $\frac{3n}{(n+1)!}$ between $0$ and $\frac{1}{n!}$? The proposed left side makes logical sense to me, however bounding the right hand side to prove the limit goes to $0$ is giving me a bit more trouble.

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    $\begingroup$ Write $3n=3(n+1)-3$. $\endgroup$ Jul 7, 2014 at 5:22

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Note that since $n<n+1$, one have that $$ 0\le \frac{3n}{(n+1)!} \le \frac{3}{n!}$$ Using Squeeze Theorem, it can be easily seen that the limit is zero.

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