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

1. geerky42

So we can see that $\prod_{n=1}^{\lfloor x\rfloor}\dfrac{x^n}{n^x}=\begin{cases} 1, & \mbox{if } x < 1 \\~\\ x, & \mbox{if } 1\le x < 2 \\~\\ \dfrac{x^3}{2^x}, & \mbox{if } 2\le x<3 \\~\\ \dfrac{x^6}{6^x}, & \mbox{if } 3\le x<4 \\~\\ \dfrac{x^{10}}{24^x}, & \mbox{if } 4\le x<5 \\ \vdots \\~\\ \Large \dfrac{x^{\frac{k(k+1)}{2}}}{(k!)^x}, & \mbox{if } k\le x<k+1 \\ \vdots \end{cases},\quad\quad k\in\mathbb Z$ But we cannot go to last case we can evaluate with... So how else can I evaluate this limit?

2. jagr2713

*CRYS*

3. geerky42

|dw:1433808236581:dw|

4. geerky42

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

5. jagr2713

Idk lol

6. yolomcswagginsggg

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

7. geerky42

I know, I was not talking to you lol @jagr2713 Anyone knows? @welshfella @jim_thompson5910 @Zarkon

8. geerky42

@yolomcswagginsggg |dw:1433808486316:dw|

9. yolomcswagginsggg

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

10. anonymous

there

11. anonymous

oh ok

12. geerky42

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

13. geerky42

Any lucks? @freckles

14. anonymous

weird

15. anonymous

need help

16. geerky42

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

17. freckles

I'm trying to do squeeze theorem

18. freckles

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

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

20. freckles

k is going to infinity because x is

21. freckles

but I'm stuck now

22. geerky42

Ah yes squeeze theorem is great. Still we have ugly limits to evaluate lol. I still have trouble with it.

23. geerky42

dang

24. geerky42

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

25. freckles

we can try to do that

26. freckles

made a type-o have to start over

27. geerky42

I am working on applying an exponential of logarithm of expression. $$a\Leftrightarrow e^{\ln a}$$

28. freckles

I will try some more later Peace for now.

29. geerky42

Ok thanks for your times.

30. geerky42

Can you try? @SithsAndGiggles

31. geerky42

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

32. anonymous

My first impression is, "Stirling approximation?"

33. 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.

34. geerky42

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

35. anonymous

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

36. 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...

37. geerky42

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

38. 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.

39. anonymous

i need help

40. 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 $$\lim\limits_{t\to\infty}e^{-t}=0$$.

41. anonymous

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

42. 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*}