Why $ lim_nrightarrow infty fracn!n^k(n-k)! =1 $? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Limits involving factorials $lim_Ntoinfty fracN!(N-k)!N^k$Finding limit of sequence: $lim _n to infty fracn!n^k(n-k)!=1$Why does $lim_ntoinfty fracn!(n-k)!n^k$ equal 1Proof that $limlimits_h to infty frach!h^k(h-k)!=1 $ for any $ k $Evaluating $lim_nrightarrowinftyleft(1-fracxn^1+aright)^n$Computing $lim_nrightarrowinfty(1-fracxn)^-n$Dominated convergence theorem for complex-valued functions?Limit to Expectation: $ - lim_N rightarrow infty frac1N sum_n=1^N fracpartialpartial theta ln p(x_n|theta)$Evaluation of $lim_mtoinftyBig(F(e^-fraclambdam^2)Big)^m$ given $F(z)=frac1-sqrt1-z^2z$Compute $lim_nrightarrowinftyleft(fracn+1nright)^n^2cdotfrac1e^n.$Find $lim_nrightarrow inftyfrac(2n-1)!!(2n)!!.$Finding $limsup_nrightarrowinfty n^fraclog(n)n$How to prove $lim_n rightarrowinfty e^-nsum_k=0^nfracn^kk! = frac12$?Show: $lim_nrightarrow infty left|int_1^eleft[ln(x)right]^n:dx right|= 0 $
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Why $ lim_nrightarrow infty fracn!n^k(n-k)! =1 $?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Limits involving factorials $lim_Ntoinfty fracN!(N-k)!N^k$Finding limit of sequence: $lim _n to infty fracn!n^k(n-k)!=1$Why does $lim_ntoinfty fracn!(n-k)!n^k$ equal 1Proof that $limlimits_h to infty frach!h^k(h-k)!=1 $ for any $ k $Evaluating $lim_nrightarrowinftyleft(1-fracxn^1+aright)^n$Computing $lim_nrightarrowinfty(1-fracxn)^-n$Dominated convergence theorem for complex-valued functions?Limit to Expectation: $ - lim_N rightarrow infty frac1N sum_n=1^N fracpartialpartial theta ln p(x_n|theta)$Evaluation of $lim_mtoinftyBig(F(e^-fraclambdam^2)Big)^m$ given $F(z)=frac1-sqrt1-z^2z$Compute $lim_nrightarrowinftyleft(fracn+1nright)^n^2cdotfrac1e^n.$Find $lim_nrightarrow inftyfrac(2n-1)!!(2n)!!}.$Finding $limsup_nrightarrowinfty n^{fraclog(n)n$How to prove $lim_n rightarrowinfty e^-nsum_k=0^nfracn^kk! = frac12$?Show: $lim_nrightarrow infty left|int_1^eleft[ln(x)right]^n:dx right|= 0 $
$begingroup$
I was on brilliant.org learning probability. There was a process explaining how the distribution of a Poisson Random Variable can be obtained from a Binomial Random Variable.
Consider the binomial distribution:
$$
beginequationbeginaligned
P(X=k) &=binom n k p^k (1-p)^n-k\
&=fracn!k!(n-k)! p^k (1-p)^n-k
endalignedendequation
$$
Substitute $m=np$ , or $p=fracmn$ :
$$
beginequationbeginaligned
P(X=k) &=fracn!k!(n-k)! left(fracmnright)^k left(1-fracmnright)^n-k\
&=fracn!k!(n-k)! fracm^kn^k left(1-fracmnright)^nleft(1-fracmnright)^-k
endalignedendequation
$$
Slightly rearrange
$$
beginequationbeginaligned
&=fracn!n^k(n-k)! left(1-fracmnright)^-kfracm^kk!left(1-fracmnright)^n
endalignedendequation
$$
Note that
$$
beginequationbeginaligned
& lim_nrightarrow infty fracn!n^k(n-k)! =1,quadlim_nrightarrow infty left(1-fracmnright)^-k =1,quad lim_nrightarrow infty left(1-fracmnright)^n =e^-m
endalignedendequation
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=fracm^k e^-mk!
$$
In all these steps, what I don't understand is the following limit:
$$
lim_nrightarrow infty fracn!n^k(n-k)! =1
$$
limits factorial
asked Apr 13 at 4:32
billyandrbillyandr
237
$endgroup$
add a comment |
$begingroup$
I was on brilliant.org learning probability. There was a process explaining how the distribution of a Poisson Random Variable can be obtained from a Binomial Random Variable.
Consider the binomial distribution:
$$
beginequationbeginaligned
P(X=k) &=binom n k p^k (1-p)^n-k\
&=fracn!k!(n-k)! p^k (1-p)^n-k
endalignedendequation
$$
Substitute $m=np$ , or $p=fracmn$ :
$$
beginequationbeginaligned
P(X=k) &=fracn!k!(n-k)! left(fracmnright)^k left(1-fracmnright)^n-k\
&=fracn!k!(n-k)! fracm^kn^k left(1-fracmnright)^nleft(1-fracmnright)^-k
endalignedendequation
$$
Slightly rearrange
$$
beginequationbeginaligned
&=fracn!n^k(n-k)! left(1-fracmnright)^-kfracm^kk!left(1-fracmnright)^n
endalignedendequation
$$
Note that
$$
beginequationbeginaligned
& lim_nrightarrow infty fracn!n^k(n-k)! =1,quadlim_nrightarrow infty left(1-fracmnright)^-k =1,quad lim_nrightarrow infty left(1-fracmnright)^n =e^-m
endalignedendequation
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=fracm^k e^-mk!
$$
In all these steps, what I don't understand is the following limit:
$$
lim_nrightarrow infty fracn!n^k(n-k)! =1
$$
limits factorial
asked Apr 13 at 4:32
billyandrbillyandr
237
$endgroup$
$begingroup$
There are several posts about this: Why does $lim_ntoinfty fracn!(n-k)!n^k$ equal 1, Finding limit of sequence: $lim _n to infty fracn!n^k(n-k)!=1$, Proof that $limlimits_h to infty frach!h^k(h-k)!=1$ for any $k$, Limits involing Factorials $lim_Ntoinfty fracN!(N-k)!N^k$
$endgroup$
– Martin Sleziak
Apr 13 at 7:46
$begingroup$
I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
$endgroup$
– Martin Sleziak
Apr 13 at 7:51
$begingroup$
Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
$endgroup$
– billyandr
Apr 13 at 23:13
$begingroup$
billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
$endgroup$
– Martin Sleziak
Apr 13 at 23:17
add a comment |
$begingroup$
I was on brilliant.org learning probability. There was a process explaining how the distribution of a Poisson Random Variable can be obtained from a Binomial Random Variable.
Consider the binomial distribution:
$$
beginequationbeginaligned
P(X=k) &=binom n k p^k (1-p)^n-k\
&=fracn!k!(n-k)! p^k (1-p)^n-k
endalignedendequation
$$
Substitute $m=np$ , or $p=fracmn$ :
$$
beginequationbeginaligned
P(X=k) &=fracn!k!(n-k)! left(fracmnright)^k left(1-fracmnright)^n-k\
&=fracn!k!(n-k)! fracm^kn^k left(1-fracmnright)^nleft(1-fracmnright)^-k
endalignedendequation
$$
Slightly rearrange
$$
beginequationbeginaligned
&=fracn!n^k(n-k)! left(1-fracmnright)^-kfracm^kk!left(1-fracmnright)^n
endalignedendequation
$$
Note that
$$
beginequationbeginaligned
& lim_nrightarrow infty fracn!n^k(n-k)! =1,quadlim_nrightarrow infty left(1-fracmnright)^-k =1,quad lim_nrightarrow infty left(1-fracmnright)^n =e^-m
endalignedendequation
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=fracm^k e^-mk!
$$
In all these steps, what I don't understand is the following limit:
$$
lim_nrightarrow infty fracn!n^k(n-k)! =1
$$
limits factorial
asked Apr 13 at 4:32
billyandrbillyandr
237
$endgroup$
I was on brilliant.org learning probability. There was a process explaining how the distribution of a Poisson Random Variable can be obtained from a Binomial Random Variable.
Consider the binomial distribution:
$$
beginequationbeginaligned
P(X=k) &=binom n k p^k (1-p)^n-k\
&=fracn!k!(n-k)! p^k (1-p)^n-k
endalignedendequation
$$
Substitute $m=np$ , or $p=fracmn$ :
$$
beginequationbeginaligned
P(X=k) &=fracn!k!(n-k)! left(fracmnright)^k left(1-fracmnright)^n-k\
&=fracn!k!(n-k)! fracm^kn^k left(1-fracmnright)^nleft(1-fracmnright)^-k
endalignedendequation
$$
Slightly rearrange
$$
beginequationbeginaligned
&=fracn!n^k(n-k)! left(1-fracmnright)^-kfracm^kk!left(1-fracmnright)^n
endalignedendequation
$$
Note that
$$
beginequationbeginaligned
& lim_nrightarrow infty fracn!n^k(n-k)! =1,quadlim_nrightarrow infty left(1-fracmnright)^-k =1,quad lim_nrightarrow infty left(1-fracmnright)^n =e^-m
endalignedendequation
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=fracm^k e^-mk!
$$
In all these steps, what I don't understand is the following limit:
$$
lim_nrightarrow infty fracn!n^k(n-k)! =1
$$
limits factorial
limits factorial
asked Apr 13 at 4:32
billyandrbillyandr
237
asked Apr 13 at 4:32
billyandrbillyandr
237
edited Apr 13 at 15:39
billyandr
asked Apr 13 at 4:32
billyandrbillyandr
237
asked Apr 13 at 4:32
billyandrbillyandr
237
asked Apr 13 at 4:32
billyandrbillyandr
237
237
$begingroup$
There are several posts about this: Why does $lim_ntoinfty fracn!(n-k)!n^k$ equal 1, Finding limit of sequence: $lim _n to infty fracn!n^k(n-k)!=1$, Proof that $limlimits_h to infty frach!h^k(h-k)!=1$ for any $k$, Limits involing Factorials $lim_Ntoinfty fracN!(N-k)!N^k$
$endgroup$
– Martin Sleziak
Apr 13 at 7:46
$begingroup$
I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
$endgroup$
– Martin Sleziak
Apr 13 at 7:51
$begingroup$
Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
$endgroup$
– billyandr
Apr 13 at 23:13
$begingroup$
billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
$endgroup$
– Martin Sleziak
Apr 13 at 23:17
add a comment |
$begingroup$
There are several posts about this: Why does $lim_ntoinfty fracn!(n-k)!n^k$ equal 1, Finding limit of sequence: $lim _n to infty fracn!n^k(n-k)!=1$, Proof that $limlimits_h to infty frach!h^k(h-k)!=1$ for any $k$, Limits involing Factorials $lim_Ntoinfty fracN!(N-k)!N^k$
$endgroup$
– Martin Sleziak
Apr 13 at 7:46
$begingroup$
I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
$endgroup$
– Martin Sleziak
Apr 13 at 7:51
$begingroup$
Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
$endgroup$
– billyandr
Apr 13 at 23:13
$begingroup$
billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
$endgroup$
– Martin Sleziak
Apr 13 at 23:17
$begingroup$
There are several posts about this: Why does $lim_ntoinfty fracn!(n-k)!n^k$ equal 1, Finding limit of sequence: $lim _n to infty fracn!n^k(n-k)!=1$, Proof that $limlimits_h to infty frach!h^k(h-k)!=1$ for any $k$, Limits involing Factorials $lim_Ntoinfty fracN!(N-k)!N^k$
$endgroup$
– Martin Sleziak
Apr 13 at 7:46
$begingroup$
There are several posts about this: Why does $lim_ntoinfty fracn!(n-k)!n^k$ equal 1, Finding limit of sequence: $lim _n to infty fracn!n^k(n-k)!=1$, Proof that $limlimits_h to infty frach!h^k(h-k)!=1$ for any $k$, Limits involing Factorials $lim_Ntoinfty fracN!(N-k)!N^k$
$endgroup$
– Martin Sleziak
Apr 13 at 7:46
$begingroup$
I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
$endgroup$
– Martin Sleziak
Apr 13 at 7:51
$begingroup$
I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
$endgroup$
– Martin Sleziak
Apr 13 at 7:51
$begingroup$
Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
$endgroup$
– billyandr
Apr 13 at 23:13
$begingroup$
Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
$endgroup$
– billyandr
Apr 13 at 23:13
$begingroup$
billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
$endgroup$
– Martin Sleziak
Apr 13 at 23:17
$begingroup$
billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
$endgroup$
– Martin Sleziak
Apr 13 at 23:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is rather obvious if you cancel the factorials:
$$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$
answered Apr 13 at 4:59
trancelocationtrancelocation
14.6k1929
$endgroup$
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
Apr 13 at 5:01
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
Apr 13 at 5:03
add a comment |
$begingroup$
$$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
$$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$
answered Apr 13 at 5:03
Claude LeiboviciClaude Leibovici
126k1158134
$endgroup$
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
Apr 13 at 5:06
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
Apr 13 at 5:11
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is rather obvious if you cancel the factorials:
$$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$
answered Apr 13 at 4:59
trancelocationtrancelocation
14.6k1929
$endgroup$
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
Apr 13 at 5:01
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
Apr 13 at 5:03
add a comment |
$begingroup$
It is rather obvious if you cancel the factorials:
$$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$
answered Apr 13 at 4:59
trancelocationtrancelocation
14.6k1929
$endgroup$
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
Apr 13 at 5:01
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
Apr 13 at 5:03
add a comment |
$begingroup$
It is rather obvious if you cancel the factorials:
$$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$
answered Apr 13 at 4:59
trancelocationtrancelocation
14.6k1929
$endgroup$
It is rather obvious if you cancel the factorials:
$$fracn!n^k(n-k)! =fracoverbracen(n-1)cdots (n-k+1)^k; factorsn^k= 1cdot left(1-frac1nright)cdots left(1-frack-1nright)stackreln to inftylongrightarrow 1$$
answered Apr 13 at 4:59
trancelocationtrancelocation
14.6k1929
answered Apr 13 at 4:59
trancelocationtrancelocation
14.6k1929
answered Apr 13 at 4:59
trancelocationtrancelocation
14.6k1929
answered Apr 13 at 4:59
trancelocationtrancelocation
14.6k1929
14.6k1929
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
Apr 13 at 5:01
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
Apr 13 at 5:03
add a comment |
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
Apr 13 at 5:01
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
Apr 13 at 5:03
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
Apr 13 at 5:01
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
Apr 13 at 5:01
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
Apr 13 at 5:03
$begingroup$
You are welcome. This "not seeing the obvious" just happens once in a while, I think, to all who do maths. So, it is good to have a math platform like this one. :-)
$endgroup$
– trancelocation
Apr 13 at 5:03
add a comment |
$begingroup$
$$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
$$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$
answered Apr 13 at 5:03
Claude LeiboviciClaude Leibovici
126k1158134
$endgroup$
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
Apr 13 at 5:06
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
Apr 13 at 5:11
add a comment |
$begingroup$
$$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
$$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$
answered Apr 13 at 5:03
Claude LeiboviciClaude Leibovici
126k1158134
$endgroup$
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
Apr 13 at 5:06
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
Apr 13 at 5:11
add a comment |
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$$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
$$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$
answered Apr 13 at 5:03
Claude LeiboviciClaude Leibovici
126k1158134
$endgroup$
$$a_n=fracn!n^k(n-k)! implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frack(1-k)2 n+Oleft(frac1n^2right)$$ Continue with Taylor
$$a_n=e^log(a_n)=1+frack(1-k)2 n+Oleft(frac1n^2right)$$
answered Apr 13 at 5:03
Claude LeiboviciClaude Leibovici
126k1158134
answered Apr 13 at 5:03
Claude LeiboviciClaude Leibovici
126k1158134
answered Apr 13 at 5:03
Claude LeiboviciClaude Leibovici
126k1158134
answered Apr 13 at 5:03
Claude LeiboviciClaude Leibovici
126k1158134
126k1158134
1
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This has already a slight touch of overkill, hasn't it? :-)
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– trancelocation
Apr 13 at 5:06
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
Apr 13 at 5:11
add a comment |
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
Apr 13 at 5:06
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
Apr 13 at 5:11
1
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
Apr 13 at 5:06
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
Apr 13 at 5:06
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
Apr 13 at 5:11
$begingroup$
@trancelocation. You are totally right for the limit. One of my manias is to always look at the approach to the limit. Have a look at matheducators.stackexchange.com/questions/8339/… . Cheers :-)
$endgroup$
– Claude Leibovici
Apr 13 at 5:11
add a comment |
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There are several posts about this: Why does $lim_ntoinfty fracn!(n-k)!n^k$ equal 1, Finding limit of sequence: $lim _n to infty fracn!n^k(n-k)!=1$, Proof that $limlimits_h to infty frach!h^k(h-k)!=1$ for any $k$, Limits involing Factorials $lim_Ntoinfty fracN!(N-k)!N^k$
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– Martin Sleziak
Apr 13 at 7:46
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I found the posts in the above comment using Approach0. For some useful tips on searching here see: How to search on this site?
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– Martin Sleziak
Apr 13 at 7:51
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Some of the other posts treating the same question painfully lack details and context. Maybe you'd want to put them on hold or close them.
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– billyandr
Apr 13 at 23:13
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billyandr: If you actually have a look at those links, you can see that two of those posts are closed (as duplicates) now. Let me also say that the fact that you have added some more context to your question is certainly appreciated. (After all, that's what lead to reopening.)
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– Martin Sleziak
Apr 13 at 23:17