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

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

Mathematics
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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
*CRYS*
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Other answers:

I know answer is 0. But how can I SHOW that?
Idk lol
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I know, I was not talking to you lol @jagr2713 Anyone knows? @welshfella @jim_thompson5910 @Zarkon
@yolomcswagginsggg |dw:1433808486316:dw|
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there
oh ok
Enough with those things. I want to know how to approach this problem
Any lucks? @freckles
weird
need help
Not really "help," but rather I just want to know a way to show that.
I'm trying to do squeeze theorem
let me show you what I have so far and what I want to prove
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: \[\lim_{k \rightarrow \infty}\frac{(k)^\frac{k(k+1)}{2}}{(k!)^{k+1}}=0 \text{ and } \lim_{k \rightarrow \infty}\frac{(k+1)^\frac{k(k+1)}{2}}{(k!)^{k}}=0\]
k is going to infinity because x is
but I'm stuck now
Ah yes squeeze theorem is great. Still we have ugly limits to evaluate lol. I still have trouble with it.
dang
For some reason, I have a feel that we need to apply squeeze theorem AGAIN.
we can try to do that
made a type-o have to start over
I am working on applying an exponential of logarithm of expression. \(a\Leftrightarrow e^{\ln a}\)
I will try some more later Peace for now.
Ok thanks for your times.
Can you try? @SithsAndGiggles
@SithsAndGiggles We just have to show: \[\lim_{k \rightarrow \infty}\frac{(k)^\frac{k(k+1)}{2}}{(k!)^{k+1}}=0 \text{ and } \lim_{k \rightarrow \infty}\frac{(k+1)^\frac{k(k+1)}{2}}{(k!)^{k}}=0\]
My first impression is, "Stirling approximation?"
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.
Any idea what to do using "Stirling approximation"? @SithsAndGiggles
Ah but that was just a gut instinct suggestion without any actual work on my part :P
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...
Man I am tired from all of this already...
\[k!\sim \sqrt{2\pi k}\left(\frac{k}{e}\right)^k\] 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.
i need help
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 \(\lim\limits_{t\to\infty}e^{-t}=0\).
Perhaps establishing that \(x^x\ge e^x\) for all \(x\ge x^*\) will suffice.
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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