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geerky42

  • one year ago

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

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

  2. jagr2713
    • one year ago
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    *CRYS*

  3. geerky42
    • one year ago
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    |dw:1433808236581:dw|

  4. geerky42
    • one year ago
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    I know answer is 0. But how can I SHOW that?

  5. jagr2713
    • one year ago
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    Idk lol

  6. yolomcswagginsggg
    • one year ago
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    ...................__ ............./´¯/'...'/´¯¯`·¸ ........../'/.../..../......./¨¯\ ........('(...´...´.... ¯~/'...') .........\.................'...../ ..........''...\.......... _.·´ ............\..............( BROFIST ...........

  7. geerky42
    • one year ago
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    I know, I was not talking to you lol @jagr2713 Anyone knows? @welshfella @jim_thompson5910 @Zarkon

  8. geerky42
    • one year ago
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    @yolomcswagginsggg |dw:1433808486316:dw|

  9. yolomcswagginsggg
    • one year ago
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    ...................__ ............./´¯/'...'/´¯¯`·¸ ........../'/.../..../......./¨¯\ ........('(...´...´.... ¯~/'...') .........\.................'...../ ..........''...\.......... _.·´ ............\..............( BROFIST ...........

  10. anonymous
    • one year ago
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    there

  11. anonymous
    • one year ago
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    oh ok

  12. geerky42
    • one year ago
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    Enough with those things. I want to know how to approach this problem

  13. geerky42
    • one year ago
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    Any lucks? @freckles

  14. anonymous
    • one year ago
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    weird

  15. anonymous
    • one year ago
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    need help

  16. geerky42
    • one year ago
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    Not really "help," but rather I just want to know a way to show that.

  17. freckles
    • one year ago
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    I'm trying to do squeeze theorem

  18. freckles
    • one year ago
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    let me show you what I have so far and what I want to prove

  19. freckles
    • one year ago
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    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
    • one year ago
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    k is going to infinity because x is

  21. freckles
    • one year ago
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    but I'm stuck now

  22. geerky42
    • one year ago
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    Ah yes squeeze theorem is great. Still we have ugly limits to evaluate lol. I still have trouble with it.

  23. geerky42
    • one year ago
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    dang

  24. geerky42
    • one year ago
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    For some reason, I have a feel that we need to apply squeeze theorem AGAIN.

  25. freckles
    • one year ago
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    we can try to do that

  26. freckles
    • one year ago
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    made a type-o have to start over

  27. geerky42
    • one year ago
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    I am working on applying an exponential of logarithm of expression. \(a\Leftrightarrow e^{\ln a}\)

  28. freckles
    • one year ago
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    I will try some more later Peace for now.

  29. geerky42
    • one year ago
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    Ok thanks for your times.

  30. geerky42
    • one year ago
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    Can you try? @SithsAndGiggles

  31. geerky42
    • one year ago
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    @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
    • one year ago
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    My first impression is, "Stirling approximation?"

  33. geerky42
    • one year ago
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    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
    • one year ago
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    Any idea what to do using "Stirling approximation"? @SithsAndGiggles

  35. anonymous
    • one year ago
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    Ah but that was just a gut instinct suggestion without any actual work on my part :P

  36. geerky42
    • one year ago
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    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
    • one year ago
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    Man I am tired from all of this already...

  38. anonymous
    • one year ago
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    \[k!\sim \sqrt{2\pi k}\left(\frac{k}{e}\right)^k\] 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
    • one year ago
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    i need help

  40. anonymous
    • one year ago
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    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
    • one year ago
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    Perhaps establishing that \(x^x\ge e^x\) for all \(x\ge x^*\) will suffice.

  42. anonymous
    • one year ago
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    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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