Question

Possible terrifying problem:



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

Answer 1

So we can see that \[\prod_{n=1}^{\lfloor x\rfloor}\dfrac{x^n}{n^x}=\begin{cases} 1, & \mbox{if } x < 1 \\~\\ x, & \mbox{if } 1\le x < 2 \\~\\ \dfrac{x^3}{2^x}, & \mbox{if } 2\le x<3 \\~\\ \dfrac{x^6}{6^x}, & \mbox{if } 3\le x<4 \\~\\ \dfrac{x^{10}}{24^x}, & \mbox{if } 4\le x<5 \\ \vdots \\~\\ \Large \dfrac{x^{\frac{k(k+1)}{2}}}{(k!)^x}, & \mbox{if } k\le x<k+1 \\ \vdots \end{cases},\quad\quad k\in\mathbb Z\]

But we cannot go to last case we can evaluate with...

So how else can I evaluate this limit?

Answer 2

*CRYS*

Answer 3

I know answer is 0. But how can I SHOW that?

Answer 4

Idk lol

Answer 5

...................__
............./´¯/'...'/´¯¯`·¸
........../'/.../..../......./¨¯\
........('(...´...´.... ¯~/'...')
.........\.................'...../
..........''...\.......... _.·´
............\..............(
BROFIST ...........

Answer 6

I know, I was not talking to you lol @jagr2713

Anyone knows? @welshfella @jim_thompson5910 @Zarkon

Answer 7

@yolomcswagginsggg

Answer 8

...................__
............./´¯/'...'/´¯¯`·¸
........../'/.../..../......./¨¯\
........('(...´...´.... ¯~/'...')
.........\.................'...../
..........''...\.......... _.·´
............\..............(
BROFIST ...........

Answer 9

there

Answer 10

oh ok

Answer 11

Enough with those things. I want to know how to approach this problem

Answer 12

Any lucks? @freckles

Answer 13

weird

Answer 14

need help

Answer 15

Not really "help," but rather I just want to know a way to show that.

Answer 16

I'm trying to do squeeze theorem

Answer 17

let me show you what I have so far
and what I want to prove

Answer 18

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:
\[\lim_{k \rightarrow \infty}\frac{(k)^\frac{k(k+1)}{2}}{(k!)^{k+1}}=0 \text{ and } \lim_{k \rightarrow \infty}\frac{(k+1)^\frac{k(k+1)}{2}}{(k!)^{k}}=0\]

Answer 19

k is going to infinity because x is

Answer 20

but I'm stuck now

Answer 21

Ah yes squeeze theorem is great.

Still we have ugly limits to evaluate lol. I still have trouble with it.

Answer 22

dang

Answer 23

For some reason, I have a feel that we need to apply squeeze theorem AGAIN.

Answer 24

we can try to do that

Answer 25

made a type-o
have to start over

Answer 26

I am working on applying an exponential of logarithm of expression.

\(a\Leftrightarrow e^{\ln a}\)

Answer 27

I will try some more later

Peace for now.

Answer 28

Ok thanks for your times.

Answer 29

Can you try? @SithsAndGiggles

Answer 30

@SithsAndGiggles

We just have to show:
\[\lim_{k \rightarrow \infty}\frac{(k)^\frac{k(k+1)}{2}}{(k!)^{k+1}}=0 \text{ and } \lim_{k \rightarrow \infty}\frac{(k+1)^\frac{k(k+1)}{2}}{(k!)^{k}}=0\]

Answer 31

My first impression is, "Stirling approximation?"

Answer 32

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.

Answer 33

Any idea what to do using "Stirling approximation"? @SithsAndGiggles

Answer 34

Ah but that was just a gut instinct suggestion without any actual work on my part :P

Answer 35

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...

Answer 36

Man I am tired from all of this already...

Answer 37

\[k!\sim \sqrt{2\pi k}\left(\frac{k}{e}\right)^k\]
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.

Answer 38

i need help

Answer 39

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 \(\lim\limits_{t\to\infty}e^{-t}=0\).

Answer 40

Perhaps establishing that \(x^x\ge e^x\) for all \(x\ge x^*\) will suffice.

Answer 41

For all \(x>e\), you have that \(x(\ln x-1)>0\), so starting with this inequality we can directly extract the desired relationship:
\[\begin{align*}
x(\ln x-1)&>0\\
x\ln x&>x\\
e^{x\ln x}&>e^x\\
x^x&>e^x
\end{align*}\]