How would you take the limit of (n+2)!/(n+3)!? I know the limit is 0 because the bottom is bigger but how do we show the work? I am thinking about using L'Hospital's rule but I don't think it would work in this case. Please help me on this. Thank you.
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You can note that:
(n + 3)! = (n + 3) * (n + 2) * (n + 1) * n * ... * 3 * 2 * 1
= (n + 3) * [(n + 2) * (n + 1) * n * ... * 3 * 2 * 1]
= (n + 3)(n + 2)!, by the definition of the factorial.
Then, we see that:
lim (n-->infinity) (n + 2)!/(n + 3)!
= lim (n-->infinity) (n + 2)!/[(n + 3)(n + 2)!]
= lim (n-->infinity) 1/(n + 3), by canceling (n + 2)!
= 0, as required.
I hope this helps!
(n + 3)! = (n + 3) * (n + 2) * (n + 1) * n * ... * 3 * 2 * 1
= (n + 3) * [(n + 2) * (n + 1) * n * ... * 3 * 2 * 1]
= (n + 3)(n + 2)!, by the definition of the factorial.
Then, we see that:
lim (n-->infinity) (n + 2)!/(n + 3)!
= lim (n-->infinity) (n + 2)!/[(n + 3)(n + 2)!]
= lim (n-->infinity) 1/(n + 3), by canceling (n + 2)!
= 0, as required.
I hope this helps!