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**particlejohn** Prove that the sequence $\displaystyle a_{n} = \frac{n^3}{n!} $ converges.

Intuitively this seems to converge to $\displaystyle 0 $ (e.g. its $\displaystyle o(n) $). So for all $\displaystyle \varepsilon > 0 $ there is an $\displaystyle N \in \mathbb{N} $, such that for $\displaystyle n \geq N $, $\displaystyle |a_n| < \varepsilon $.

Is that the end of the proof? Would you choose an $\displaystyle N \in \mathbb{N} $ such that $\displaystyle \frac{N^3}{N!} < \varepsilon $?