I'm not trying to start an argument here but your answer isn't correct since my example contradicts it. You didn't explain anything
about polynomials, and even if you did that doesn't prove that a sequence which is given in form of a fraction will
necessarily converge to zero if simply, as you wrote, "the denominator grows faster than the numerator".
If a monotone ascending sequence has an upper bound, then it converges. If a monotone descending sequence has a lower bound, then it converges. Sequences of positive numbers have, by definition of "positive", 0 as a lower bound.
I have another question, I understand that if a sequence is monotone ascending and have an upper bound, there is a limit, and if monotone descending and have a lower bound there is a limit. And to show a sequence is ascending I have to show An < An+1 and for descending An > An+1. But for this problem n^n/(n!e^e), I know it is descending and has lower bound of 0 since it is positive, however what is the proper way of showing that An > An+1 ??? I can't think of anything else but plugging in...
Use ratio test proposed by Drexel28 above:
An/An+1= (n+1)!/n! X n^n/(n+1)^(n+1) X e^(n+1)/e^n
= (n+1/n)! X (n/n+1)^n X 1/(n+1) X e^(n+1)/e^n
= (n+1/n)^n X e > 1
An+1/An < 1 => monotonically decreaing, and > 0.
(Sorry about Cauchy. Didn't read everything and assumed ratio test had been tried.)
I'm not saying that this is wrong, but maybe a better counterexample would be something like .
I also think that dwsmith's statement is correct if we add the word "much." That is, the denominator grows "much faster" than the numerator (in the original question). When I say that a function grows much faster than a function , by that I mean . Although I suppose this is just rephrasing the original question.