Why $ lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1 $? Announcing the arrival of Valued...
Mortgage adviser recommends a longer term than necessary combined with overpayments
Can a monk deflect thrown melee weapons?
Single author papers against my advisor's will?
Unexpected result with right shift after bitwise negation
Two different pronunciation of "понял"
Classification of bundles, Postnikov towers, obstruction theory, local coefficients
Do we know why communications with Beresheet and NASA were lost during the attempted landing of the Moon lander?
Cold is to Refrigerator as warm is to?
What do you call the holes in a flute?
What's the point in a preamp?
Losing the Initialization Vector in Cipher Block Chaining
Does a C shift expression have unsigned type? Why would Splint warn about a right-shift?
Keep going mode for require-package
How to rotate it perfectly?
Autumning in love
Slither Like a Snake
How can I make names more distinctive without making them longer?
Antler Helmet: Can it work?
How do I automatically answer y in bash script?
Stop battery usage [Ubuntu 18]
Why is "Captain Marvel" translated as male in Portugal?
I'm thinking of a number
Strange behaviour of Check
How do I keep my slimes from escaping their pens?
Why $ lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1 $?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1Limits involving factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1 $ for any $ k $Evaluating $lim_{nrightarrowinfty}left(1-frac{x}{n^{1+a}}right)^{n}$Computing $lim_{nrightarrowinfty}(1-frac{x}{n})^{-n}$Dominated convergence theorem for complex-valued functions?Limit to Expectation: $ - lim_{N rightarrow infty } frac{1}{N} sum_{n=1}^{N} frac{partial}{partial theta} ln p(x_n|theta)$Evaluation of $lim_{mtoinfty}Big(F(e^{-frac{lambda}{m^2}})Big)^m$ given $F(z)=frac{1-sqrt{1-z^2}}{z}$Compute $lim_{nrightarrowinfty}left(frac{n+1}{n}right)^{n^2}cdotfrac{1}{e^n}.$Find $lim_{nrightarrow infty}frac{(2n-1)!!}{(2n)!!}.$Finding $limsup_{nrightarrowinfty} n^{frac{log(n)}{n}}$How to prove $lim_{n rightarrowinfty} e^{-n}sum_{k=0}^{n}frac{n^k}{k!} = frac{1}{2}$?Show: $lim_{nrightarrow infty} left|int_{1}^{e}left[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:
$$
begin{equation}begin{aligned}
P(X=k) &={binom n k} p^k (1-p)^{n-k}\
&=frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}
end{aligned}end{equation}
$$
Substitute $m=np$ , or $p=frac{m}{n}$ :
$$
begin{equation}begin{aligned}
P(X=k) &=frac{n!}{k!(n-k)!} left(frac{m}{n}right)^k left(1-frac{m}{n}right)^{n-k}\
&=frac{n!}{k!(n-k)!} frac{m^k}{n^k} left(1-frac{m}{n}right)^{n}left(1-frac{m}{n}right)^{-k}
end{aligned}end{equation}
$$
Slightly rearrange
$$
begin{equation}begin{aligned}
&=frac{n!}{n^k(n-k)!} left(1-frac{m}{n}right)^{-k}frac{m^k}{k!}left(1-frac{m}{n}right)^{n}
end{aligned}end{equation}
$$
Note that
$$
begin{equation}begin{aligned}
& lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1,quadlim_{nrightarrow infty} left(1-frac{m}{n}right)^{-k} =1,quad lim_{nrightarrow infty} left(1-frac{m}{n}right)^{n} =e^{-m}
end{aligned}end{equation}
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=frac{m^k e^{-m}}{k!}
$$
In all these steps, what I don't understand is the following limit:
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
limits factorial
asked yesterday
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:
$$
begin{equation}begin{aligned}
P(X=k) &={binom n k} p^k (1-p)^{n-k}\
&=frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}
end{aligned}end{equation}
$$
Substitute $m=np$ , or $p=frac{m}{n}$ :
$$
begin{equation}begin{aligned}
P(X=k) &=frac{n!}{k!(n-k)!} left(frac{m}{n}right)^k left(1-frac{m}{n}right)^{n-k}\
&=frac{n!}{k!(n-k)!} frac{m^k}{n^k} left(1-frac{m}{n}right)^{n}left(1-frac{m}{n}right)^{-k}
end{aligned}end{equation}
$$
Slightly rearrange
$$
begin{equation}begin{aligned}
&=frac{n!}{n^k(n-k)!} left(1-frac{m}{n}right)^{-k}frac{m^k}{k!}left(1-frac{m}{n}right)^{n}
end{aligned}end{equation}
$$
Note that
$$
begin{equation}begin{aligned}
& lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1,quadlim_{nrightarrow infty} left(1-frac{m}{n}right)^{-k} =1,quad lim_{nrightarrow infty} left(1-frac{m}{n}right)^{n} =e^{-m}
end{aligned}end{equation}
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=frac{m^k e^{-m}}{k!}
$$
In all these steps, what I don't understand is the following limit:
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
limits factorial
asked yesterday
billyandrbillyandr
237
$endgroup$
$begingroup$
There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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:
$$
begin{equation}begin{aligned}
P(X=k) &={binom n k} p^k (1-p)^{n-k}\
&=frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}
end{aligned}end{equation}
$$
Substitute $m=np$ , or $p=frac{m}{n}$ :
$$
begin{equation}begin{aligned}
P(X=k) &=frac{n!}{k!(n-k)!} left(frac{m}{n}right)^k left(1-frac{m}{n}right)^{n-k}\
&=frac{n!}{k!(n-k)!} frac{m^k}{n^k} left(1-frac{m}{n}right)^{n}left(1-frac{m}{n}right)^{-k}
end{aligned}end{equation}
$$
Slightly rearrange
$$
begin{equation}begin{aligned}
&=frac{n!}{n^k(n-k)!} left(1-frac{m}{n}right)^{-k}frac{m^k}{k!}left(1-frac{m}{n}right)^{n}
end{aligned}end{equation}
$$
Note that
$$
begin{equation}begin{aligned}
& lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1,quadlim_{nrightarrow infty} left(1-frac{m}{n}right)^{-k} =1,quad lim_{nrightarrow infty} left(1-frac{m}{n}right)^{n} =e^{-m}
end{aligned}end{equation}
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=frac{m^k e^{-m}}{k!}
$$
In all these steps, what I don't understand is the following limit:
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
limits factorial
asked yesterday
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:
$$
begin{equation}begin{aligned}
P(X=k) &={binom n k} p^k (1-p)^{n-k}\
&=frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}
end{aligned}end{equation}
$$
Substitute $m=np$ , or $p=frac{m}{n}$ :
$$
begin{equation}begin{aligned}
P(X=k) &=frac{n!}{k!(n-k)!} left(frac{m}{n}right)^k left(1-frac{m}{n}right)^{n-k}\
&=frac{n!}{k!(n-k)!} frac{m^k}{n^k} left(1-frac{m}{n}right)^{n}left(1-frac{m}{n}right)^{-k}
end{aligned}end{equation}
$$
Slightly rearrange
$$
begin{equation}begin{aligned}
&=frac{n!}{n^k(n-k)!} left(1-frac{m}{n}right)^{-k}frac{m^k}{k!}left(1-frac{m}{n}right)^{n}
end{aligned}end{equation}
$$
Note that
$$
begin{equation}begin{aligned}
& lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1,quadlim_{nrightarrow infty} left(1-frac{m}{n}right)^{-k} =1,quad lim_{nrightarrow infty} left(1-frac{m}{n}right)^{n} =e^{-m}
end{aligned}end{equation}
$$
Thus, we have the final result which is equal to the formula for the Poisson distribution.
$$
=frac{m^k e^{-m}}{k!}
$$
In all these steps, what I don't understand is the following limit:
$$
lim_{nrightarrow infty} frac{n!}{n^{k}(n-k)! } =1
$$
limits factorial
limits factorial
asked yesterday
billyandrbillyandr
237
asked yesterday
billyandrbillyandr
237
edited yesterday
billyandr
asked yesterday
billyandrbillyandr
237
asked yesterday
billyandrbillyandr
237
asked yesterday
billyandrbillyandr
237
237
$begingroup$
There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
add a comment |
$begingroup$
There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered yesterday
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
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
yesterday
add a comment |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@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
yesterday
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3185830%2fwhy-lim-n-rightarrow-infty-fracnnkn-k-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered yesterday
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
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
yesterday
add a comment |
$begingroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered yesterday
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
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
yesterday
add a comment |
$begingroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered yesterday
trancelocationtrancelocation
14.1k1829
$endgroup$
It is rather obvious if you cancel the factorials:
$$frac{n!}{n^{k}(n-k)! } =frac{overbrace{n(n-1)cdots (n-k+1)}^{k; factors}}{n^k}= 1cdot left(1-frac{1}{n}right)cdots left(1-frac{k-1}{n}right)stackrel{n to infty}{longrightarrow} 1$$
answered yesterday
trancelocationtrancelocation
14.1k1829
answered yesterday
trancelocationtrancelocation
14.1k1829
answered yesterday
trancelocationtrancelocation
14.1k1829
answered yesterday
trancelocationtrancelocation
14.1k1829
14.1k1829
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
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
yesterday
add a comment |
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
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
yesterday
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
Thank you so much. I didn't know it was right there under my eyes.
$endgroup$
– billyandr
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
add a comment |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@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
yesterday
add a comment |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@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
yesterday
add a comment |
$begingroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
$$a_n=frac{n!}{n^{k}(n-k)! }implies log(a_n)=log(n!)-k log(n)-log((n-k)!)$$
Use Stirling approximation and continue with Taylor series to get
$$log(a_n)=frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=1+frac{k(1-k)}{2 n}+Oleft(frac{1}{n^2}right)$$
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
answered yesterday
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@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
yesterday
add a comment |
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@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
yesterday
1
1
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
This has already a slight touch of overkill, hasn't it? :-)
$endgroup$
– trancelocation
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3185830%2fwhy-lim-n-rightarrow-infty-fracnnkn-k-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
There are several posts about this: Why does $lim_{ntoinfty} frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $lim _{n to infty} {frac{n!}{n^k(n-k)!}}=1$, Proof that $limlimits_{h to infty} frac{h!}{h^k(h-k)!}=1$ for any $k$, Limits involing Factorials $lim_{Ntoinfty} frac{N!}{(N-k)!N^{k}}$
$endgroup$
– Martin Sleziak
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday