geerky42
  • geerky42
Possible terrifying problem: Evaluate: \[\huge \lim_{x\to\infty}\prod_{n=1}^{\lfloor x\rfloor}\dfrac{x^n}{n^x}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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geerky42
  • geerky42
So 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
jagr2713
  • jagr2713
*CRYS*
geerky42
  • geerky42
|dw:1433808236581:dw|

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geerky42
  • geerky42
I know answer is 0. But how can I SHOW that?
jagr2713
  • jagr2713
Idk lol
yolomcswagginsggg
  • yolomcswagginsggg
...................__ ............./´¯/'...'/´¯¯`·¸ ........../'/.../..../......./¨¯\ ........('(...´...´.... ¯~/'...') .........\.................'...../ ..........''...\.......... _.·´ ............\..............( BROFIST ...........
geerky42
  • geerky42
I know, I was not talking to you lol @jagr2713 Anyone knows? @welshfella @jim_thompson5910 @Zarkon
geerky42
  • geerky42
@yolomcswagginsggg |dw:1433808486316:dw|
yolomcswagginsggg
  • yolomcswagginsggg
...................__ ............./´¯/'...'/´¯¯`·¸ ........../'/.../..../......./¨¯\ ........('(...´...´.... ¯~/'...') .........\.................'...../ ..........''...\.......... _.·´ ............\..............( BROFIST ...........
anonymous
  • anonymous
there
anonymous
  • anonymous
oh ok
geerky42
  • geerky42
Enough with those things. I want to know how to approach this problem
geerky42
  • geerky42
Any lucks? @freckles
anonymous
  • anonymous
weird
anonymous
  • anonymous
need help
geerky42
  • geerky42
Not really "help," but rather I just want to know a way to show that.
freckles
  • freckles
I'm trying to do squeeze theorem
freckles
  • freckles
let me show you what I have so far and what I want to prove
freckles
  • freckles
want 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
  • freckles
k is going to infinity because x is
freckles
  • freckles
but I'm stuck now
geerky42
  • geerky42
Ah yes squeeze theorem is great. Still we have ugly limits to evaluate lol. I still have trouble with it.
geerky42
  • geerky42
dang
geerky42
  • geerky42
For some reason, I have a feel that we need to apply squeeze theorem AGAIN.
freckles
  • freckles
we can try to do that
freckles
  • freckles
made a type-o have to start over
geerky42
  • geerky42
I am working on applying an exponential of logarithm of expression. \(a\Leftrightarrow e^{\ln a}\)
freckles
  • freckles
I will try some more later Peace for now.
geerky42
  • geerky42
Ok thanks for your times.
geerky42
  • geerky42
Can you try? @SithsAndGiggles
geerky42
  • geerky42
@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
  • anonymous
My first impression is, "Stirling approximation?"
geerky42
  • geerky42
Going "\(a\Leftrightarrow e^{\ln a}\)" and simplify further, I end up with \[\lim_{k\to\infty }\left(\dfrac{ke^{k/2}}{(k-1)!}\right)^{k+1}\]. Hope I didn't make mistake somewhere.
geerky42
  • geerky42
Any idea what to do using "Stirling approximation"? @SithsAndGiggles
anonymous
  • anonymous
Ah but that was just a gut instinct suggestion without any actual work on my part :P
geerky42
  • geerky42
Oh ah. Well, I am at \[\lim_{k\to\infty }\left(\dfrac{ke^{k/2}}{(k-1)!}\right)^{k+1}\] I have a feel that it should be easy to evaluate, but I am stuck...
geerky42
  • geerky42
Man I am tired from all of this already...
anonymous
  • anonymous
\[k!\sim \sqrt{2\pi k}\left(\frac{k}{e}\right)^k\] 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
  • anonymous
i need help
anonymous
  • anonymous
Some 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
  • anonymous
Perhaps establishing that \(x^x\ge e^x\) for all \(x\ge x^*\) will suffice.
anonymous
  • anonymous
For all \(x>e\), you have that \(x(\ln x-1)>0\), so starting with this inequality we can directly extract the desired relationship: \[\begin{align*} x(\ln x-1)&>0\\ x\ln x&>x\\ e^{x\ln x}&>e^x\\ x^x&>e^x \end{align*}\]

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