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About LinkBacks1. Show that lim (n^n)/(n!e^n) exists without finding the limit.
n->oo
2. Show that lim (1/n){(2*4*...*(2n))/(1*3*...*(2n-1))}^2 exists without finding the limit. n->oo
I guess I have to prove this by least upper bounds, however hard to start!
Limits of a simple polynomial we just evaluate the fraction of the highest power variable. I would have changed my answer to take n/n then but for this case and since it doesn't say solve the limit, I gave the poster a correct answer for his limit.Originally Posted by tonio
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".
Tonio
To show it is a monotone descending sequence, I just have to show that n < n+1 right?? and then how can I prove it converges???
need little more help on both of them
Well, as both sequences are obviously positive, showing they are monotonee descending automatically
makes them monotone and bounded and thus converging.
Tonio
To show that a sequence is monotone ascending you would need to show that $\displaystyle a_n< a_{n+1}$. To show a sequence is monotone descending, you would need to show that $\displaystyle a_n> a_{n+1}$.To show it is a monotone descending sequence, I just have to show that n < n+1 right?? and then how can I prove it converges???
need little more help on both of them
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...
As in my last post, if you put $\displaystyle \displaystyle{a_n:=\frac{n^n}{n!\,e^n}}$ , then we get
$\displaystyle \displaystyle{a_{n+1}\leq a_n\Longleftrightarrow \frac{(n+1)^{n+1}}{(n+1)!\,e^{n+1}}\leq \frac{n^n}{n!\,e^n}}\Longleftrightarrow \frac{(n+1)^n}{n^n}\leq e}$ , and we get
that this last inequality is true since, as was observed in my past message, the left side
in it converges monotonically ascending to the right side.
Tonio
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 wouldn't say that $\displaystyle n+1$ grows faster than $\displaystyle n$. I would say that $\displaystyle n+1$ is greater than $\displaystyle n$ for all $\displaystyle n$, but that they grow at the same rate.
I'm not saying that this is wrong, but maybe a better counterexample would be something like $\displaystyle \frac{n^2}{n^2+n}$.
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 $\displaystyle f$ grows much faster than a function $\displaystyle g$, by that I mean $\displaystyle \lim_{x\rightarrow \infty\frac{g(x)}{f(x)}=0$. Although I suppose this is just rephrasing the original question.
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