Limit with Factorial: Understanding 3/(n+1) Limit

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Homework Statement


Why does the limit as n -> infinity of [3^(n+1)]/(n+1)!] * n!/(3^n) equal
the limit as n -> infinity of 3/(n+1)?

Homework Equations


The Attempt at a Solution


I have never encountered this before.
 
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Forget about the limit and just focus on simplifying algebraically. You know that {(n+1)}! = n!{(n+1)}. Also 3^{n+1} = (3)(3^n). Use those to cancel some terms and see what you get.
 
Forget about the limit, how do you simplify:

[3^(n+1)]/(n+1)! * [n!/(3^n)]

(LOL, Curious3141 is faster than me)
 
yenchin said:
Forget about the limit, how do you simplify:

[3^(n+1)]/(n+1)! * [n!/(3^n)]

(LOL, Curious3141 is faster than me)

And it's weird how we worded that almost identically! :eek:
 
Ah yes, that makes perfect sense. Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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