1. 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!

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- Jan 20th 2011, 05:35 PMmathsohardLimits
1. 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! - Jan 20th 2011, 05:38 PMdwsmith
- Jan 20th 2011, 06:56 PMtonio
- Jan 20th 2011, 07:00 PMdwsmith
- Jan 20th 2011, 07:12 PMtonio
- Jan 20th 2011, 07:16 PMtonio

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 - Jan 20th 2011, 08:10 PMDrexel28
- Jan 20th 2011, 10:59 PMmathsohard
- Jan 21st 2011, 05:50 AMtonio
- Jan 21st 2011, 06:04 AMHallsofIvy
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 $\displaystyle a_n> a_{n+1}$.

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. - Jan 21st 2011, 12:12 PMmathsohard
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...

- Jan 21st 2011, 02:16 PMtonio

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 - Jan 22nd 2011, 11:31 AMHartlwCauchy Convergence Theorem
It seems to me if you have to show something converges without finding a limit then Cauchy's convergence theorem is worth a try.

- Jan 24th 2011, 03:50 AMHartlw
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.) - Jan 24th 2011, 04:29 AMDrSteve
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.