$lim_x to infty (x!)^pover x^x$ The Next CEO of Stack OverflowEvaluate $lim_ntoinftyprod_k=1^nfrac2k2k-1$How can I find if $sum_n=1^infty n! over 10^n $ converges or diverges?How to find $lim_nto inftyfracn^n-1n!$Prove that $lim_uto inftyfracu^me^u=0$Determine $sumlimits_n=1^infty frac1(n!)!$ conv or div?What is $lim_nto infty sqrt[n+1](n+1)! over sqrt[n]n!$Limit $lim_x to inftyx^e/e^x$Determining: $lim_nrightarrow inftye^-nsum_k=n^infty n^k/k!$Evaluate $sum_r=1^infty sqrt frac rr^4+r^2+1$Calculating $lim_nto inftyfracfracn1 + fracn-12 + fracn-23 + … + frac2n-1 + frac1nln(n!)$
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$lim_x to infty (x!)^pover x^x$
The Next CEO of Stack OverflowEvaluate $lim_ntoinftyprod_k=1^nfrac2k2k-1$How can I find if $sum_n=1^infty n! over 10^n $ converges or diverges?How to find $lim_nto inftyfracn^n-1n!$Prove that $lim_uto inftyfracu^me^u=0$Determine $sumlimits_n=1^infty frac1(n!)!$ conv or div?What is $lim_nto infty sqrt[n+1](n+1)! over sqrt[n]n!$Limit $lim_x to inftyx^e/e^x$Determining: $lim_nrightarrow inftye^-nsum_k=n^infty n^k/k!$Evaluate $sum_r=1^infty} sqrt {frac rr^4+r^2+1$Calculating $lim_nto inftyfracfracn1 + fracn-12 + fracn-23 + … + frac2n-1 + frac1nln(n!)$
$begingroup$
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
asked Mar 26 at 14:23
William GrannisWilliam Grannis
1,030521
$endgroup$
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
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
asked Mar 26 at 14:23
William GrannisWilliam Grannis
1,030521
$endgroup$
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
calculus limits factorial
asked Mar 26 at 14:23
William GrannisWilliam Grannis
1,030521
asked Mar 26 at 14:23
William GrannisWilliam Grannis
1,030521
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
1,030521
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 |
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
1
1
$begingroup$
I assume you mean $ple 1$ vs $p> 1$.
$endgroup$
– J.G.
Mar 26 at 14:25
$begingroup$
I assume you mean $ple 1$ vs $p> 1$.
$endgroup$
– J.G.
Mar 26 at 14:25
1
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
$begingroup$
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 Answers
2
active
oldest
votes
$begingroup$
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 $ln xin o(x)$.
answered Mar 26 at 14:30
J.G.J.G.
32.3k23250
$endgroup$
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.
answered 2 days ago
William GrannisWilliam Grannis
1,030521
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $ln xin o(x)$.
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 $ln xin o(x)$.
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 $ln xin o(x)$.
answered Mar 26 at 14:30
J.G.J.G.
32.3k23250
$endgroup$
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 $ln xin o(x)$.
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
answered Mar 26 at 14:30
J.G.J.G.
32.3k23250
32.3k23250
add a comment |
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.
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.
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.
answered 2 days ago
William GrannisWilliam Grannis
1,030521
$endgroup$
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.
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
answered 2 days ago
William GrannisWilliam Grannis
1,030521
1,030521
add a comment |
add a comment |
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$begingroup$
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