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.

  • 2
    $\begingroup$ Write $3n=3(n+1)-3$. $\endgroup$ Jul 7, 2014 at 5:22

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .