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$lim_x to infty (x!)^pover x^x$

2

$begingroup$

After some experimentation, I am pretty sure that $$lim_x to infty (x!)^pover x^x to infty $$ for $p gt 1$, and that
$$lim_x to infty (x!)^pover x^x to 0 $$ for $p le 1$.

I do not, however, have any proof of this, and I was wondering if someone could help me. I have tried applying the ratio test, but I am not quite sure how. Thanks in advance for the help!

calculus limits factorial

share|cite|improve this question

edited 2 days ago

William Grannis

asked Mar 26 at 14:23

William GrannisWilliam Grannis

1,030521

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I assume you mean $ple 1$ vs $p> 1$.
  $endgroup$
  – J.G.
  Mar 26 at 14:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Please edit your question. First of all, you probably want to separate the cases based on $p$ and not on $x$. Second, the expression is larger if $p>1$, so the limit should also be larger...
  $endgroup$
  – 5xum
  Mar 26 at 14:29
  

  
  
  
  

add a comment | 

2

$begingroup$

After some experimentation, I am pretty sure that $$lim_x to infty (x!)^pover x^x to infty $$ for $p gt 1$, and that
$$lim_x to infty (x!)^pover x^x to 0 $$ for $p le 1$.

I do not, however, have any proof of this, and I was wondering if someone could help me. I have tried applying the ratio test, but I am not quite sure how. Thanks in advance for the help!

calculus limits factorial

share|cite|improve this question

edited 2 days ago

William Grannis

asked Mar 26 at 14:23

William GrannisWilliam Grannis

1,030521

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I assume you mean $ple 1$ vs $p> 1$.
  $endgroup$
  – J.G.
  Mar 26 at 14:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Please edit your question. First of all, you probably want to separate the cases based on $p$ and not on $x$. Second, the expression is larger if $p>1$, so the limit should also be larger...
  $endgroup$
  – 5xum
  Mar 26 at 14:29
  

  
  
  
  

add a comment | 

$begingroup$

After some experimentation, I am pretty sure that $$lim_x to infty (x!)^pover x^x to infty $$ for $p gt 1$, and that
$$lim_x to infty (x!)^pover x^x to 0 $$ for $p le 1$.

I do not, however, have any proof of this, and I was wondering if someone could help me. I have tried applying the ratio test, but I am not quite sure how. Thanks in advance for the help!

calculus limits factorial

share|cite|improve this question

edited 2 days ago

William Grannis

asked Mar 26 at 14:23

William GrannisWilliam Grannis

1,030521

$endgroup$

share|cite|improve this question

edited 2 days ago

William Grannis

asked Mar 26 at 14:23

William GrannisWilliam Grannis

1,030521

share|cite|improve this question

edited 2 days ago

William Grannis

asked Mar 26 at 14:23

William GrannisWilliam Grannis

1,030521

asked Mar 26 at 14:23

William GrannisWilliam Grannis

1,030521

asked Mar 26 at 14:23

William GrannisWilliam Grannis

1,030521

2 Answers
2

active

oldest

votes

4

$begingroup$

The Stirling approximation states $x!simsqrt2pix^x+1/2e^-x$. Thus $$lim_xtoinftyfracx!^px^x=(2pi)^p/2explim_xtoinftyleft[left((p-1)x+fracp2right)ln x-pxright].$$If $p>1$, this is $expinfty=infty$ due to the $xln x$ term. If $p<1$, the same logic obtains a limit of $exp-infty=0$. If $p=1$, we get $$sqrt2piexplim_xtoinfty(tfrac12ln x-x)=exp-infty=0$$ because $ln xin o(x)$.

share|cite|improve this answer

answered Mar 26 at 14:30

J.G.J.G.

32.3k23250

$endgroup$

add a comment | 

1

$begingroup$

Solution via Ratio Test:

The ratio between two successive terms is: $$(n!)^povern^nover((n+1)!)^pover(n+1)^n+1$$

Simplifying gives:
$$left(n!over(n+1)!right)^p cdot (n+1)^n+1over n^n$$

Further simplification yields:
$$1over (n+1)^p cdot left(n+1over nright)^ncdot (n+1)$$

With some rearrangement we have:
$$1over (n+1)^p-1 cdot left(n+1over nright)^n$$

The second term goes off to $e$ as $n$ goes to infinity. The first goes to $0$ when $p$ is larger than $1$, $1$ when $p$ is equal to $1$, and $+infty$ when $p$ is less than $1$. In the first case, the ratio between successive terms is $0$. In the second case, it is $e$, and in the third case it is unbounded. By the ratio test, it must convege in the first case and diverge in the other two.

share|cite|improve this answer

answered 2 days ago

William GrannisWilliam Grannis

1,030521

$endgroup$

add a comment | 

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4

$begingroup$

The Stirling approximation states $x!simsqrt2pix^x+1/2e^-x$. Thus $$lim_xtoinftyfracx!^px^x=(2pi)^p/2explim_xtoinftyleft[left((p-1)x+fracp2right)ln x-pxright].$$If $p>1$, this is $expinfty=infty$ due to the $xln x$ term. If $p<1$, the same logic obtains a limit of $exp-infty=0$. If $p=1$, we get $$sqrt2piexplim_xtoinfty(tfrac12ln x-x)=exp-infty=0$$ because $ln xin o(x)$.

share|cite|improve this answer

answered Mar 26 at 14:30

J.G.J.G.

32.3k23250

$endgroup$

add a comment | 

4

$begingroup$

The Stirling approximation states $x!simsqrt2pix^x+1/2e^-x$. Thus $$lim_xtoinftyfracx!^px^x=(2pi)^p/2explim_xtoinftyleft[left((p-1)x+fracp2right)ln x-pxright].$$If $p>1$, this is $expinfty=infty$ due to the $xln x$ term. If $p<1$, the same logic obtains a limit of $exp-infty=0$. If $p=1$, we get $$sqrt2piexplim_xtoinfty(tfrac12ln x-x)=exp-infty=0$$ because $ln xin o(x)$.

share|cite|improve this answer

answered Mar 26 at 14:30

J.G.J.G.

32.3k23250

$endgroup$

add a comment | 

$begingroup$

The Stirling approximation states $x!simsqrt2pix^x+1/2e^-x$. Thus $$lim_xtoinftyfracx!^px^x=(2pi)^p/2explim_xtoinftyleft[left((p-1)x+fracp2right)ln x-pxright].$$If $p>1$, this is $expinfty=infty$ due to the $xln x$ term. If $p<1$, the same logic obtains a limit of $exp-infty=0$. If $p=1$, we get $$sqrt2piexplim_xtoinfty(tfrac12ln x-x)=exp-infty=0$$ because $ln xin o(x)$.

share|cite|improve this answer

answered Mar 26 at 14:30

J.G.J.G.

32.3k23250

$endgroup$

share|cite|improve this answer

answered Mar 26 at 14:30

J.G.J.G.

32.3k23250

answered Mar 26 at 14:30

J.G.J.G.

32.3k23250

answered Mar 26 at 14:30

J.G.J.G.

32.3k23250

1

$begingroup$

Solution via Ratio Test:

The ratio between two successive terms is: $$(n!)^povern^nover((n+1)!)^pover(n+1)^n+1$$

Simplifying gives:
$$left(n!over(n+1)!right)^p cdot (n+1)^n+1over n^n$$

Further simplification yields:
$$1over (n+1)^p cdot left(n+1over nright)^ncdot (n+1)$$

With some rearrangement we have:
$$1over (n+1)^p-1 cdot left(n+1over nright)^n$$

The second term goes off to $e$ as $n$ goes to infinity. The first goes to $0$ when $p$ is larger than $1$, $1$ when $p$ is equal to $1$, and $+infty$ when $p$ is less than $1$. In the first case, the ratio between successive terms is $0$. In the second case, it is $e$, and in the third case it is unbounded. By the ratio test, it must convege in the first case and diverge in the other two.

share|cite|improve this answer

answered 2 days ago

William GrannisWilliam Grannis

1,030521

$endgroup$

add a comment | 

1

$begingroup$

Solution via Ratio Test:

The ratio between two successive terms is: $$(n!)^povern^nover((n+1)!)^pover(n+1)^n+1$$

Simplifying gives:
$$left(n!over(n+1)!right)^p cdot (n+1)^n+1over n^n$$

Further simplification yields:
$$1over (n+1)^p cdot left(n+1over nright)^ncdot (n+1)$$

With some rearrangement we have:
$$1over (n+1)^p-1 cdot left(n+1over nright)^n$$

The second term goes off to $e$ as $n$ goes to infinity. The first goes to $0$ when $p$ is larger than $1$, $1$ when $p$ is equal to $1$, and $+infty$ when $p$ is less than $1$. In the first case, the ratio between successive terms is $0$. In the second case, it is $e$, and in the third case it is unbounded. By the ratio test, it must convege in the first case and diverge in the other two.

share|cite|improve this answer

answered 2 days ago

William GrannisWilliam Grannis

1,030521

$endgroup$

add a comment | 

$begingroup$

Solution via Ratio Test:

The ratio between two successive terms is: $$(n!)^povern^nover((n+1)!)^pover(n+1)^n+1$$

Simplifying gives:
$$left(n!over(n+1)!right)^p cdot (n+1)^n+1over n^n$$

Further simplification yields:
$$1over (n+1)^p cdot left(n+1over nright)^ncdot (n+1)$$

With some rearrangement we have:
$$1over (n+1)^p-1 cdot left(n+1over nright)^n$$

The second term goes off to $e$ as $n$ goes to infinity. The first goes to $0$ when $p$ is larger than $1$, $1$ when $p$ is equal to $1$, and $+infty$ when $p$ is less than $1$. In the first case, the ratio between successive terms is $0$. In the second case, it is $e$, and in the third case it is unbounded. By the ratio test, it must convege in the first case and diverge in the other two.

share|cite|improve this answer

answered 2 days ago

William GrannisWilliam Grannis

1,030521

$endgroup$

share|cite|improve this answer

answered 2 days ago

William GrannisWilliam Grannis

1,030521

answered 2 days ago

William GrannisWilliam Grannis

1,030521

answered 2 days ago

William GrannisWilliam Grannis

1,030521