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geerky42
 one year ago
Possible terrifying problem:
Evaluate: \[\huge \lim_{x\to\infty}\prod_{n=1}^{\lfloor x\rfloor}\dfrac{x^n}{n^x}\]
geerky42
 one year ago
Possible terrifying problem: Evaluate: \[\huge \lim_{x\to\infty}\prod_{n=1}^{\lfloor x\rfloor}\dfrac{x^n}{n^x}\]

This Question is Closed

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6So we can see that \[\prod_{n=1}^{\lfloor x\rfloor}\dfrac{x^n}{n^x}=\begin{cases} 1, & \mbox{if } x < 1 \\~\\ x, & \mbox{if } 1\le x < 2 \\~\\ \dfrac{x^3}{2^x}, & \mbox{if } 2\le x<3 \\~\\ \dfrac{x^6}{6^x}, & \mbox{if } 3\le x<4 \\~\\ \dfrac{x^{10}}{24^x}, & \mbox{if } 4\le x<5 \\ \vdots \\~\\ \Large \dfrac{x^{\frac{k(k+1)}{2}}}{(k!)^x}, & \mbox{if } k\le x<k+1 \\ \vdots \end{cases},\quad\quad k\in\mathbb Z\] But we cannot go to last case we can evaluate with... So how else can I evaluate this limit?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6dw:1433808236581:dw

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6I know answer is 0. But how can I SHOW that?

yolomcswagginsggg
 one year ago
Best ResponseYou've already chosen the best response.0...................__ ............./´¯/'...'/´¯¯`·¸ ........../'/.../..../......./¨¯\ ........('(...´...´.... ¯~/'...') .........\.................'...../ ..........''...\.......... _.·´ ............\..............( BROFIST ...........

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6I know, I was not talking to you lol @jagr2713 Anyone knows? @welshfella @jim_thompson5910 @Zarkon

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6@yolomcswagginsggg dw:1433808486316:dw

yolomcswagginsggg
 one year ago
Best ResponseYou've already chosen the best response.0...................__ ............./´¯/'...'/´¯¯`·¸ ........../'/.../..../......./¨¯\ ........('(...´...´.... ¯~/'...') .........\.................'...../ ..........''...\.......... _.·´ ............\..............( BROFIST ...........

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6Enough with those things. I want to know how to approach this problem

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6Not really "help," but rather I just want to know a way to show that.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to do squeeze theorem

freckles
 one year ago
Best ResponseYou've already chosen the best response.0let me show you what I have so far and what I want to prove

freckles
 one year ago
Best ResponseYou've already chosen the best response.0want to prove \[\lim_{x \rightarrow \infty} \frac{x^{\frac{k(k+1)}{2}}}{(k!)^x } =0 \] where x is in [k,k+1) and where k>0 \[k \le x < k+1 \\ (k)^{\frac{k(k+1)}{2}} \le x^\frac{k(k+1)}{2} < (k+1)^{\frac{k(k+1)}{2}} \\ \\ \frac{ (k)^{ \frac{k(k+1)}{2}}}{(k!)^{k+1}} \le \frac{x^\frac{k(k+1)}{2}}{(k!)^x} < \frac{(k+1)^{\frac{k(k+1)}{2}} }{(k!)^k}\] anyways we have that we need to show: \[\lim_{k \rightarrow \infty}\frac{(k)^\frac{k(k+1)}{2}}{(k!)^{k+1}}=0 \text{ and } \lim_{k \rightarrow \infty}\frac{(k+1)^\frac{k(k+1)}{2}}{(k!)^{k}}=0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0k is going to infinity because x is

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6Ah yes squeeze theorem is great. Still we have ugly limits to evaluate lol. I still have trouble with it.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6For some reason, I have a feel that we need to apply squeeze theorem AGAIN.

freckles
 one year ago
Best ResponseYou've already chosen the best response.0we can try to do that

freckles
 one year ago
Best ResponseYou've already chosen the best response.0made a typeo have to start over

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6I am working on applying an exponential of logarithm of expression. \(a\Leftrightarrow e^{\ln a}\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.0I will try some more later Peace for now.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6Ok thanks for your times.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6Can you try? @SithsAndGiggles

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6@SithsAndGiggles We just have to show: \[\lim_{k \rightarrow \infty}\frac{(k)^\frac{k(k+1)}{2}}{(k!)^{k+1}}=0 \text{ and } \lim_{k \rightarrow \infty}\frac{(k+1)^\frac{k(k+1)}{2}}{(k!)^{k}}=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My first impression is, "Stirling approximation?"

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6Going "\(a\Leftrightarrow e^{\ln a}\)" and simplify further, I end up with \[\lim_{k\to\infty }\left(\dfrac{ke^{k/2}}{(k1)!}\right)^{k+1}\]. Hope I didn't make mistake somewhere.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6Any idea what to do using "Stirling approximation"? @SithsAndGiggles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah but that was just a gut instinct suggestion without any actual work on my part :P

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6Oh ah. Well, I am at \[\lim_{k\to\infty }\left(\dfrac{ke^{k/2}}{(k1)!}\right)^{k+1}\] I have a feel that it should be easy to evaluate, but I am stuck...

geerky42
 one year ago
Best ResponseYou've already chosen the best response.6Man I am tired from all of this already...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[k!\sim \sqrt{2\pi k}\left(\frac{k}{e}\right)^k\] so \[\frac{k^{k(k+1)/2}}{(k!)^{k+1}}=\frac{k^{(k+1)/2}}{(2\pi)^{(k+1)/2}e^{k(k+1)}k^{(k+1)/2+k(k+1)}}\] The \(k\) factors will simplify: \[\frac{k+1}{2}\left(\frac{k+1}{2}+k(k+1)\right)=k(k+1)\] giving \[\frac{k^{k(k+1)/2}}{(k!)^{k+1}}=\frac{e^{k(k+1)}}{(2\pi)^{(k+1)/2}k^{k(k+1)}}\] This might be easier to work with.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Some rearrangement allows us to work with the following limit: \[\lim_{k\to\infty}\left(\frac{e^k}{\sqrt{2\pi}k^k}\right)^{k+1}\] Taking exponentials and logarithms gives \[\exp\left(\lim_{k\to\infty}(k+1)\ln\frac{e^k}{\sqrt{2\pi}k^k}\right)\] I don't think this next observation is a valid one, so I'm looking for another way to consider this, but as \(k\to\infty\) you have \(k+1\to\infty\) and \(\dfrac{e^k}{k^k}\to0\), so \(\ln\dfrac{e^k}{\sqrt{2\pi}k^k}\to\infty\). This might suggest we have an expression here that approaches \(\lim\limits_{t\to\infty}e^{t}=0\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Perhaps establishing that \(x^x\ge e^x\) for all \(x\ge x^*\) will suffice.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For all \(x>e\), you have that \(x(\ln x1)>0\), so starting with this inequality we can directly extract the desired relationship: \[\begin{align*} x(\ln x1)&>0\\ x\ln x&>x\\ e^{x\ln x}&>e^x\\ x^x&>e^x \end{align*}\]
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