[Calculus] Sequence Limits: n -> infinity (n/n^n)(Use Sandwich Rule?)

raaznar
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Homework Statement


Use sandwich Rule to find the limit lim n> infinity (a_n) of the sequences, for which the nth term, a_n, is given.


Homework Equations


^{lim}_{n\rightarrow∞}\frac{n!}{n^{n}}


The Attempt at a Solution


I know by just looking at it, n^n Approaches infinity much faster than n! which results in limit approaching 0, which is the answer. But the question says to use Sandwich Rule? I don't know which 2 functions to use to bound n!/n^n between? Usually if there was a sin function, I could start with it between -1 and 1. But I don't know where to start for this question?
 
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raaznar said:

Homework Statement


Use sandwich Rule to find the limit lim n> infinity (a_n) of the sequences, for which the nth term, a_n, is given.

Homework Equations


^{lim}_{n\rightarrow∞}\frac{n!}{n^{n}}

The Attempt at a Solution


I know by just looking at it, n^n Approaches infinity much faster than n! which results in limit approaching 0, which is the answer. But the question says to use Sandwich Rule? I don't know which 2 functions to use to bound n!/n^n between? Usually if there was a sin function, I could start with it between -1 and 1. But I don't know where to start for this question?


Expand it. n!/n^n=(n/n)*((n-1)/n)*((n-2)/n)*...*(3/n)*(2/n)*(1/n). Does that give you any ideas?
 
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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