Mixing
Limit of a quotient of factorials
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I want to find the limit of $$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!$$ for an arbitrary integer $k$. I have done several simulations for $k=5$ or $k=3$, the limit is zero in these cases.
I have tried to proof this, but the $k-th$ power gives me a headache. Is there any way to show this rigorously?
real-analysis sequences-and-series limits
asked Aug 2 at 21:46
quallenjäger
462419
add a comment |Â
up vote
2
down vote
favorite
I want to find the limit of $$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!$$ for an arbitrary integer $k$. I have done several simulations for $k=5$ or $k=3$, the limit is zero in these cases.
I have tried to proof this, but the $k-th$ power gives me a headache. Is there any way to show this rigorously?
real-analysis sequences-and-series limits
asked Aug 2 at 21:46
quallenjäger
462419
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I want to find the limit of $$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!$$ for an arbitrary integer $k$. I have done several simulations for $k=5$ or $k=3$, the limit is zero in these cases.
I have tried to proof this, but the $k-th$ power gives me a headache. Is there any way to show this rigorously?
real-analysis sequences-and-series limits
asked Aug 2 at 21:46
quallenjäger
462419
I want to find the limit of $$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!$$ for an arbitrary integer $k$. I have done several simulations for $k=5$ or $k=3$, the limit is zero in these cases.
I have tried to proof this, but the $k-th$ power gives me a headache. Is there any way to show this rigorously?
real-analysis sequences-and-series limits
asked Aug 2 at 21:46
quallenjäger
462419
asked Aug 2 at 21:46
quallenjäger
462419
asked Aug 2 at 21:46
quallenjäger
462419
asked Aug 2 at 21:46
quallenjäger
462419
462419
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
For $k=1$ we have
$$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!=lim_ntoinftyfrac(2n)!(2n)!=1$$
and for $kge 2$ by ratio test we have
$$frac((2n+2)!)^k((2n+2)k+k-1)!frac(2nk+k-1)!((2n)!)^k=frac(2n+2)^k(2n+1)^k(2nk+3k-1)ldots(2nk+k)<$$
$$<frac(2n+2)^2k(2nk+k)^2kto k^-2k<1$$
therefore
$$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!=0$$
answered Aug 2 at 22:02
gimusi
63.8k73480
2
As $ntoinfty$, the last expression converges to $k^-2k$.
– Julián Aguirre
Aug 2 at 22:05
@JuliánAguirre Yes of course, I had in mind the final result for the limit. I fix that point! Thanks
– gimusi
Aug 2 at 22:07
@gimusi Thank you, that's a clever proof.
– quallenjäger
Aug 2 at 22:10
@quallenjäger You are welcome! Bye...and thanks again to Julian for pointing out my fault!
– gimusi
Aug 2 at 22:11
add a comment |Â
up vote
2
down vote
Considering $$A=frac((2n)!)^k(2nk+k-1)!implies log(A)=k log((2n)!)-log((2nk+k-1)!)$$ Now, use Stirling approximation
$$log(p!)=p (log (p)-1)+frac12 left(log (2 pi )+log
left(pright)right)+frac112 p+Oleft(frac1p^3right)$$
Apply it and continue with Taylor expansions to get
$$log(A)=-2 n (k log (k))-frac12 left(log left(kright)+(k-1) left(2
log (k)-log left(fracpi nright)right)right)+Oleft(frac1nright)$$ So, basically
$$A sim k^-2 k n$$
Edit
Following your question in comments, the next level of approximation would be
$$A=pi ^frack-12 k^-2 k n left(frac1k^2 nright)^k/2 sqrtk n+Oleft(frac1nright)$$
Uisng $k=10$ and $n=20$, the first approximation gives $10^-400$ while the new one gives $approx 7.63times 10^-414$ for an exact value $approx 6.97times 10^-414$
answered Aug 3 at 4:05
Claude Leibovici
111k1054126
So the quotient converges with a speed of $k^-2kn$? If I mutiple the quotient with $k^-2kn$, then the limit of the quotient will not be zero?
– quallenjäger
Aug 3 at 9:32
@quallenjäger. It is more complex than that. I suppose that you wanted to divide the result by $k^-2kn$. Let me see if I can find a better approximation.
– Claude Leibovici
Aug 3 at 10:15
Yes sorry, I meant to multiply to avoid the limit being zero.
– quallenjäger
Aug 3 at 10:16
add a comment |Â
up vote
0
down vote
$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!
$
Let's start with a simpler,
larger quotient
using Stirling
in the form
$n!
approx sqrtcn(n/e)^n$.
$beginarray\
dfrac((2n)!)^k(2nk)!
&approx dfrac(sqrt2cn(2n/e)^2n)^ksqrt2cnk(2nk/e)^2nk\
&= dfrac(2cn)^k/2(2n/e)^2nksqrt2cnk(2nk/e)^2nk\
&= dfrac(2cn)^k/2(2n)^2nk/e^2nksqrt2cnk(2nk)^2nk/e^2nk\
&= dfrac(2cn)^k/2sqrt2cnk(k)^2nk\
&= dfrac1sqrt2cnkleft(dfrac(2cn)^1/2k^2nright)^k\
&= dfrac1sqrt2cnkleft(dfrac(2cn)^1/(2n)k^2right)^kn\
&= dfrac1sqrt2cnkleft(dfrace^ln(2cn)/(2n)k^2right)^kn\
&< dfrac1sqrt2cnkleft(dfrac1+ln(2cn)/(2n)k^2right)^kn
qquadtextsince e^x le 1+2xtext for 0 le x le 1\
&to 0
qquadtextfor k gt 1\
endarray
$
answered Aug 2 at 22:30
marty cohen
69k446122
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
For $k=1$ we have
$$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!=lim_ntoinftyfrac(2n)!(2n)!=1$$
and for $kge 2$ by ratio test we have
$$frac((2n+2)!)^k((2n+2)k+k-1)!frac(2nk+k-1)!((2n)!)^k=frac(2n+2)^k(2n+1)^k(2nk+3k-1)ldots(2nk+k)<$$
$$<frac(2n+2)^2k(2nk+k)^2kto k^-2k<1$$
therefore
$$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!=0$$
answered Aug 2 at 22:02
gimusi
63.8k73480
2
As $ntoinfty$, the last expression converges to $k^-2k$.
– Julián Aguirre
Aug 2 at 22:05
@JuliánAguirre Yes of course, I had in mind the final result for the limit. I fix that point! Thanks
– gimusi
Aug 2 at 22:07
@gimusi Thank you, that's a clever proof.
– quallenjäger
Aug 2 at 22:10
@quallenjäger You are welcome! Bye...and thanks again to Julian for pointing out my fault!
– gimusi
Aug 2 at 22:11
add a comment |Â
up vote
4
down vote
accepted
For $k=1$ we have
$$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!=lim_ntoinftyfrac(2n)!(2n)!=1$$
and for $kge 2$ by ratio test we have
$$frac((2n+2)!)^k((2n+2)k+k-1)!frac(2nk+k-1)!((2n)!)^k=frac(2n+2)^k(2n+1)^k(2nk+3k-1)ldots(2nk+k)<$$
$$<frac(2n+2)^2k(2nk+k)^2kto k^-2k<1$$
therefore
$$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!=0$$
answered Aug 2 at 22:02
gimusi
63.8k73480
2
As $ntoinfty$, the last expression converges to $k^-2k$.
– Julián Aguirre
Aug 2 at 22:05
@JuliánAguirre Yes of course, I had in mind the final result for the limit. I fix that point! Thanks
– gimusi
Aug 2 at 22:07
@gimusi Thank you, that's a clever proof.
– quallenjäger
Aug 2 at 22:10
@quallenjäger You are welcome! Bye...and thanks again to Julian for pointing out my fault!
– gimusi
Aug 2 at 22:11
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
For $k=1$ we have
$$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!=lim_ntoinftyfrac(2n)!(2n)!=1$$
and for $kge 2$ by ratio test we have
$$frac((2n+2)!)^k((2n+2)k+k-1)!frac(2nk+k-1)!((2n)!)^k=frac(2n+2)^k(2n+1)^k(2nk+3k-1)ldots(2nk+k)<$$
$$<frac(2n+2)^2k(2nk+k)^2kto k^-2k<1$$
therefore
$$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!=0$$
answered Aug 2 at 22:02
gimusi
63.8k73480
For $k=1$ we have
$$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!=lim_ntoinftyfrac(2n)!(2n)!=1$$
and for $kge 2$ by ratio test we have
$$frac((2n+2)!)^k((2n+2)k+k-1)!frac(2nk+k-1)!((2n)!)^k=frac(2n+2)^k(2n+1)^k(2nk+3k-1)ldots(2nk+k)<$$
$$<frac(2n+2)^2k(2nk+k)^2kto k^-2k<1$$
therefore
$$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!=0$$
answered Aug 2 at 22:02
gimusi
63.8k73480
edited Aug 2 at 22:06
answered Aug 2 at 22:02
gimusi
63.8k73480
answered Aug 2 at 22:02
gimusi
63.8k73480
answered Aug 2 at 22:02
gimusi
63.8k73480
63.8k73480
2
As $ntoinfty$, the last expression converges to $k^-2k$.
– Julián Aguirre
Aug 2 at 22:05
@JuliánAguirre Yes of course, I had in mind the final result for the limit. I fix that point! Thanks
– gimusi
Aug 2 at 22:07
@gimusi Thank you, that's a clever proof.
– quallenjäger
Aug 2 at 22:10
@quallenjäger You are welcome! Bye...and thanks again to Julian for pointing out my fault!
– gimusi
Aug 2 at 22:11
add a comment |Â
2
As $ntoinfty$, the last expression converges to $k^-2k$.
– Julián Aguirre
Aug 2 at 22:05
@JuliánAguirre Yes of course, I had in mind the final result for the limit. I fix that point! Thanks
– gimusi
Aug 2 at 22:07
@gimusi Thank you, that's a clever proof.
– quallenjäger
Aug 2 at 22:10
@quallenjäger You are welcome! Bye...and thanks again to Julian for pointing out my fault!
– gimusi
Aug 2 at 22:11
2
2
As $ntoinfty$, the last expression converges to $k^-2k$.
– Julián Aguirre
Aug 2 at 22:05
As $ntoinfty$, the last expression converges to $k^-2k$.
– Julián Aguirre
Aug 2 at 22:05
@JuliánAguirre Yes of course, I had in mind the final result for the limit. I fix that point! Thanks
– gimusi
Aug 2 at 22:07
@JuliánAguirre Yes of course, I had in mind the final result for the limit. I fix that point! Thanks
– gimusi
Aug 2 at 22:07
@gimusi Thank you, that's a clever proof.
– quallenjäger
Aug 2 at 22:10
@gimusi Thank you, that's a clever proof.
– quallenjäger
Aug 2 at 22:10
@quallenjäger You are welcome! Bye...and thanks again to Julian for pointing out my fault!
– gimusi
Aug 2 at 22:11
@quallenjäger You are welcome! Bye...and thanks again to Julian for pointing out my fault!
– gimusi
Aug 2 at 22:11
add a comment |Â
up vote
2
down vote
Considering $$A=frac((2n)!)^k(2nk+k-1)!implies log(A)=k log((2n)!)-log((2nk+k-1)!)$$ Now, use Stirling approximation
$$log(p!)=p (log (p)-1)+frac12 left(log (2 pi )+log
left(pright)right)+frac112 p+Oleft(frac1p^3right)$$
Apply it and continue with Taylor expansions to get
$$log(A)=-2 n (k log (k))-frac12 left(log left(kright)+(k-1) left(2
log (k)-log left(fracpi nright)right)right)+Oleft(frac1nright)$$ So, basically
$$A sim k^-2 k n$$
Edit
Following your question in comments, the next level of approximation would be
$$A=pi ^frack-12 k^-2 k n left(frac1k^2 nright)^k/2 sqrtk n+Oleft(frac1nright)$$
Uisng $k=10$ and $n=20$, the first approximation gives $10^-400$ while the new one gives $approx 7.63times 10^-414$ for an exact value $approx 6.97times 10^-414$
answered Aug 3 at 4:05
Claude Leibovici
111k1054126
So the quotient converges with a speed of $k^-2kn$? If I mutiple the quotient with $k^-2kn$, then the limit of the quotient will not be zero?
– quallenjäger
Aug 3 at 9:32
@quallenjäger. It is more complex than that. I suppose that you wanted to divide the result by $k^-2kn$. Let me see if I can find a better approximation.
– Claude Leibovici
Aug 3 at 10:15
Yes sorry, I meant to multiply to avoid the limit being zero.
– quallenjäger
Aug 3 at 10:16
add a comment |Â
up vote
2
down vote
Considering $$A=frac((2n)!)^k(2nk+k-1)!implies log(A)=k log((2n)!)-log((2nk+k-1)!)$$ Now, use Stirling approximation
$$log(p!)=p (log (p)-1)+frac12 left(log (2 pi )+log
left(pright)right)+frac112 p+Oleft(frac1p^3right)$$
Apply it and continue with Taylor expansions to get
$$log(A)=-2 n (k log (k))-frac12 left(log left(kright)+(k-1) left(2
log (k)-log left(fracpi nright)right)right)+Oleft(frac1nright)$$ So, basically
$$A sim k^-2 k n$$
Edit
Following your question in comments, the next level of approximation would be
$$A=pi ^frack-12 k^-2 k n left(frac1k^2 nright)^k/2 sqrtk n+Oleft(frac1nright)$$
Uisng $k=10$ and $n=20$, the first approximation gives $10^-400$ while the new one gives $approx 7.63times 10^-414$ for an exact value $approx 6.97times 10^-414$
answered Aug 3 at 4:05
Claude Leibovici
111k1054126
So the quotient converges with a speed of $k^-2kn$? If I mutiple the quotient with $k^-2kn$, then the limit of the quotient will not be zero?
– quallenjäger
Aug 3 at 9:32
@quallenjäger. It is more complex than that. I suppose that you wanted to divide the result by $k^-2kn$. Let me see if I can find a better approximation.
– Claude Leibovici
Aug 3 at 10:15
Yes sorry, I meant to multiply to avoid the limit being zero.
– quallenjäger
Aug 3 at 10:16
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Considering $$A=frac((2n)!)^k(2nk+k-1)!implies log(A)=k log((2n)!)-log((2nk+k-1)!)$$ Now, use Stirling approximation
$$log(p!)=p (log (p)-1)+frac12 left(log (2 pi )+log
left(pright)right)+frac112 p+Oleft(frac1p^3right)$$
Apply it and continue with Taylor expansions to get
$$log(A)=-2 n (k log (k))-frac12 left(log left(kright)+(k-1) left(2
log (k)-log left(fracpi nright)right)right)+Oleft(frac1nright)$$ So, basically
$$A sim k^-2 k n$$
Edit
Following your question in comments, the next level of approximation would be
$$A=pi ^frack-12 k^-2 k n left(frac1k^2 nright)^k/2 sqrtk n+Oleft(frac1nright)$$
Uisng $k=10$ and $n=20$, the first approximation gives $10^-400$ while the new one gives $approx 7.63times 10^-414$ for an exact value $approx 6.97times 10^-414$
answered Aug 3 at 4:05
Claude Leibovici
111k1054126
Considering $$A=frac((2n)!)^k(2nk+k-1)!implies log(A)=k log((2n)!)-log((2nk+k-1)!)$$ Now, use Stirling approximation
$$log(p!)=p (log (p)-1)+frac12 left(log (2 pi )+log
left(pright)right)+frac112 p+Oleft(frac1p^3right)$$
Apply it and continue with Taylor expansions to get
$$log(A)=-2 n (k log (k))-frac12 left(log left(kright)+(k-1) left(2
log (k)-log left(fracpi nright)right)right)+Oleft(frac1nright)$$ So, basically
$$A sim k^-2 k n$$
Edit
Following your question in comments, the next level of approximation would be
$$A=pi ^frack-12 k^-2 k n left(frac1k^2 nright)^k/2 sqrtk n+Oleft(frac1nright)$$
Uisng $k=10$ and $n=20$, the first approximation gives $10^-400$ while the new one gives $approx 7.63times 10^-414$ for an exact value $approx 6.97times 10^-414$
answered Aug 3 at 4:05
Claude Leibovici
111k1054126
edited Aug 3 at 10:39
answered Aug 3 at 4:05
Claude Leibovici
111k1054126
answered Aug 3 at 4:05
Claude Leibovici
111k1054126
answered Aug 3 at 4:05
Claude Leibovici
111k1054126
111k1054126
So the quotient converges with a speed of $k^-2kn$? If I mutiple the quotient with $k^-2kn$, then the limit of the quotient will not be zero?
– quallenjäger
Aug 3 at 9:32
@quallenjäger. It is more complex than that. I suppose that you wanted to divide the result by $k^-2kn$. Let me see if I can find a better approximation.
– Claude Leibovici
Aug 3 at 10:15
Yes sorry, I meant to multiply to avoid the limit being zero.
– quallenjäger
Aug 3 at 10:16
add a comment |Â
So the quotient converges with a speed of $k^-2kn$? If I mutiple the quotient with $k^-2kn$, then the limit of the quotient will not be zero?
– quallenjäger
Aug 3 at 9:32
@quallenjäger. It is more complex than that. I suppose that you wanted to divide the result by $k^-2kn$. Let me see if I can find a better approximation.
– Claude Leibovici
Aug 3 at 10:15
Yes sorry, I meant to multiply to avoid the limit being zero.
– quallenjäger
Aug 3 at 10:16
So the quotient converges with a speed of $k^-2kn$? If I mutiple the quotient with $k^-2kn$, then the limit of the quotient will not be zero?
– quallenjäger
Aug 3 at 9:32
So the quotient converges with a speed of $k^-2kn$? If I mutiple the quotient with $k^-2kn$, then the limit of the quotient will not be zero?
– quallenjäger
Aug 3 at 9:32
@quallenjäger. It is more complex than that. I suppose that you wanted to divide the result by $k^-2kn$. Let me see if I can find a better approximation.
– Claude Leibovici
Aug 3 at 10:15
@quallenjäger. It is more complex than that. I suppose that you wanted to divide the result by $k^-2kn$. Let me see if I can find a better approximation.
– Claude Leibovici
Aug 3 at 10:15
Yes sorry, I meant to multiply to avoid the limit being zero.
– quallenjäger
Aug 3 at 10:16
Yes sorry, I meant to multiply to avoid the limit being zero.
– quallenjäger
Aug 3 at 10:16
add a comment |Â
up vote
0
down vote
$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!
$
Let's start with a simpler,
larger quotient
using Stirling
in the form
$n!
approx sqrtcn(n/e)^n$.
$beginarray\
dfrac((2n)!)^k(2nk)!
&approx dfrac(sqrt2cn(2n/e)^2n)^ksqrt2cnk(2nk/e)^2nk\
&= dfrac(2cn)^k/2(2n/e)^2nksqrt2cnk(2nk/e)^2nk\
&= dfrac(2cn)^k/2(2n)^2nk/e^2nksqrt2cnk(2nk)^2nk/e^2nk\
&= dfrac(2cn)^k/2sqrt2cnk(k)^2nk\
&= dfrac1sqrt2cnkleft(dfrac(2cn)^1/2k^2nright)^k\
&= dfrac1sqrt2cnkleft(dfrac(2cn)^1/(2n)k^2right)^kn\
&= dfrac1sqrt2cnkleft(dfrace^ln(2cn)/(2n)k^2right)^kn\
&< dfrac1sqrt2cnkleft(dfrac1+ln(2cn)/(2n)k^2right)^kn
qquadtextsince e^x le 1+2xtext for 0 le x le 1\
&to 0
qquadtextfor k gt 1\
endarray
$
answered Aug 2 at 22:30
marty cohen
69k446122
add a comment |Â
up vote
0
down vote
$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!
$
Let's start with a simpler,
larger quotient
using Stirling
in the form
$n!
approx sqrtcn(n/e)^n$.
$beginarray\
dfrac((2n)!)^k(2nk)!
&approx dfrac(sqrt2cn(2n/e)^2n)^ksqrt2cnk(2nk/e)^2nk\
&= dfrac(2cn)^k/2(2n/e)^2nksqrt2cnk(2nk/e)^2nk\
&= dfrac(2cn)^k/2(2n)^2nk/e^2nksqrt2cnk(2nk)^2nk/e^2nk\
&= dfrac(2cn)^k/2sqrt2cnk(k)^2nk\
&= dfrac1sqrt2cnkleft(dfrac(2cn)^1/2k^2nright)^k\
&= dfrac1sqrt2cnkleft(dfrac(2cn)^1/(2n)k^2right)^kn\
&= dfrac1sqrt2cnkleft(dfrace^ln(2cn)/(2n)k^2right)^kn\
&< dfrac1sqrt2cnkleft(dfrac1+ln(2cn)/(2n)k^2right)^kn
qquadtextsince e^x le 1+2xtext for 0 le x le 1\
&to 0
qquadtextfor k gt 1\
endarray
$
answered Aug 2 at 22:30
marty cohen
69k446122
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!
$
Let's start with a simpler,
larger quotient
using Stirling
in the form
$n!
approx sqrtcn(n/e)^n$.
$beginarray\
dfrac((2n)!)^k(2nk)!
&approx dfrac(sqrt2cn(2n/e)^2n)^ksqrt2cnk(2nk/e)^2nk\
&= dfrac(2cn)^k/2(2n/e)^2nksqrt2cnk(2nk/e)^2nk\
&= dfrac(2cn)^k/2(2n)^2nk/e^2nksqrt2cnk(2nk)^2nk/e^2nk\
&= dfrac(2cn)^k/2sqrt2cnk(k)^2nk\
&= dfrac1sqrt2cnkleft(dfrac(2cn)^1/2k^2nright)^k\
&= dfrac1sqrt2cnkleft(dfrac(2cn)^1/(2n)k^2right)^kn\
&= dfrac1sqrt2cnkleft(dfrace^ln(2cn)/(2n)k^2right)^kn\
&< dfrac1sqrt2cnkleft(dfrac1+ln(2cn)/(2n)k^2right)^kn
qquadtextsince e^x le 1+2xtext for 0 le x le 1\
&to 0
qquadtextfor k gt 1\
endarray
$
answered Aug 2 at 22:30
marty cohen
69k446122
$lim_ntoinftyfrac((2n)!)^k(2nk+k-1)!
$
Let's start with a simpler,
larger quotient
using Stirling
in the form
$n!
approx sqrtcn(n/e)^n$.
$beginarray\
dfrac((2n)!)^k(2nk)!
&approx dfrac(sqrt2cn(2n/e)^2n)^ksqrt2cnk(2nk/e)^2nk\
&= dfrac(2cn)^k/2(2n/e)^2nksqrt2cnk(2nk/e)^2nk\
&= dfrac(2cn)^k/2(2n)^2nk/e^2nksqrt2cnk(2nk)^2nk/e^2nk\
&= dfrac(2cn)^k/2sqrt2cnk(k)^2nk\
&= dfrac1sqrt2cnkleft(dfrac(2cn)^1/2k^2nright)^k\
&= dfrac1sqrt2cnkleft(dfrac(2cn)^1/(2n)k^2right)^kn\
&= dfrac1sqrt2cnkleft(dfrace^ln(2cn)/(2n)k^2right)^kn\
&< dfrac1sqrt2cnkleft(dfrac1+ln(2cn)/(2n)k^2right)^kn
qquadtextsince e^x le 1+2xtext for 0 le x le 1\
&to 0
qquadtextfor k gt 1\
endarray
$
answered Aug 2 at 22:30
marty cohen
69k446122
answered Aug 2 at 22:30
marty cohen
69k446122
answered Aug 2 at 22:30
marty cohen
69k446122
answered Aug 2 at 22:30
marty cohen
69k446122
69k446122
add a comment |Â
add a comment |Â
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