Originally Posted by

**Arron** Is this correct?

The dominant term is n!, so we write

$\displaystyle {a}_{n } = \frac{{n}^{4 }+2(n!) }{{n}^{3 }+{(-1)}^{n+1 }(n!) }$ = $\displaystyle \frac{{n}^{4 }/n! +2 }{{n}^{3 }/n! +{(-1)}^{n+1 } } $

Since {$\displaystyle {n}^{4 }/n! $},{$\displaystyle {n}^{3 }/n! $} and { $\displaystyle {(-1)}^{n+1 } $} are all null sequences,

$\displaystyle \lim_{n \to \infty} {a}_{n } = \lim_{n \to \infty} \frac{{n}^{4 }/n! +2 }{{n}^{3 }/n! +{(-1)}^{n+1 } } $ = $\displaystyle \frac{0+2}{0+0 } $ = 0

by the Combination Rules.

Hence $\displaystyle {a}_{n } $ converges with a limit 0