Find the limit of the sequence (if it exists). $\alpha _ { n } = \frac { ( n + 6 ) ! } { n ! }$
A) $\lim_ { n \rightarrow \infty }\alpha _ { n } = - 6$
B) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 0$
C) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 1$
D) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 6$
E) $\lim _ { n \rightarrow \infty }a _ { n } = \infty$

Question

Find the limit of the sequence (if it exists).   $\alpha _ { n } = \frac { ( n + 6 ) ! } { n ! }$  
A) $\lim_ { n \rightarrow \infty }\alpha _ { n } = - 6$
B) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 0$
C) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 1$
D) $\lim_ { n \rightarrow \infty } \alpha _ { n } = 6$
E) $\lim _ { n \rightarrow \infty }a _ { n } = \infty$

Quizplus · Accepted Answer

The sequence $\alpha _ { n } = \frac { ( n + 6 ) ! } { n ! }$ simplifies to $(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)$, which grows without bound as $n$ approaches infinity, hence the limit is $\infty$.

Find the Limit of the Sequence (If It Exists) αn=(n+6)!n!\alpha _ { n } = \frac { ( n + 6 ) ! } { n ! }αn​=n!(n+6)!​

Find the Limit of the Sequence (If It Exists) $\alpha _ { n } = \frac { ( n + 6 ) ! } { n ! }$