Mixing
Why is this not true $lim_{nto infty}sqrt[n]{(n!)} = 1$ ? Root Test For $frac{(n!)^2}{(2n)!}$.
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I have to conclude if the following problem is convergent or not using the root test or ratio test for positive series.
$$frac{(n!)^2}{(2n)!}$$
I've shown it to be convergent using the ratio test, but I've run into a problem when using the root test. I've concluded that $sqrt[n]{(n!)} to 1$ but Wolfram Alpha claims that $sqrt[n]{(n!)} to infty$.
Here is my work:
I started by reducing the expression:
$$frac{(n!)^2}{(2n)!} = frac{(n!)(n!)}{(2n)!}.$$
We have that
$$(n!) = (n)(n-1)(n-2)cdot...cdot 1 $$
$$(2n)! = (2n)(2n-1)cdot...cdot 1 $$
$$ = 2(n)(n-1/2)(n-1)(n-3/2)(n-2)cdot...cdot 1 $$ we rearrange this
$$ 2(n)(n-1)(n-2)(n-1/2)(n-3/2)cdot ...cdot 1 .$$ It should be clear that this "contains" $(n!).$ So now we have:
$$frac{(n!)(n!)}{(2n)!} = frac{(n!)}{(2n-1)(2n-3)cdot ...cdot 1 }. $$ I now applied the root test to this:
$$sqrt[n]{frac{(n!)}{(2n-1)(2-3)cdot ...cdot 1}} = {frac{sqrt[n]{(n!)}}{sqrt[n]{(2n-1)(2-3)cdot ...cdot 1}}}. $$
I examined $sqrt[n]{(n)}$:
$$sqrt[n]{(n)} = e^{(ln(n)cdot(1/n))}. $$ $e^x$ is continuous so we examine
$$lim_{nto infty}(frac{ln(n)}{n}) to frac{infty}{infty}. $$
We use L'hopital:
$$lim_{nto infty}(frac{ln(n)'}{n'}) = lim_{nto infty}(frac{1/n}{1}) = lim_{nto infty}1/n to 0.$$
So we have
$$lim_{nto infty}sqrt[n]{(n!)} = e^0 = 1.$$
We will have the same result given $sqrt[n]{(n-k)}.$
We can now return to $sqrt[n]{(n!)}$:
$$sqrt[n]{(n!)} = sqrt[n]{(n)}sqrt[n]{(n-1)}sqrt[n]{(n-2)}*..*1$$
$$lim_{nto infty}sqrt[n]{(n!)} = 1cdot1cdot1cdot ... cdot 1 = 1 .$$ I end up concluding the same thing regarding $$lim_{nto infty}sqrt[n]{{(2n-1)(2-3)cdot ... cdot 1}}.$$
I can't figure out why $sqrt[n]{(n!)} to infty,$ according to Wolfram Alpha. Is it because it becomes an indeterminant form of $1^infty$ ? Or have I simply made a mistake so that it never becomes $1cdot 1cdot 1cdot ... cdot 1,$ repeating infinitely?
I've looked at some of the other answers concerning the question of why $sqrt[n]{(n!)} to infty$ it doesn't answer why my specific argument is incorrect, and most of them seem to be using a "trick". Given what is in the book I'm using, these seem needlessly complicated or foreign to the reader. I am simply doing this for fun, the problem is already solved by using the ratio test, so perhaps I am simply underestimating the complexity of this problem.
real-analysis sequences-and-series limits
edited Jan 21 at 0:10
jordan_glen
1
asked Jan 20 at 22:22
DimajoDimajo
42
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|
show 2 more comments
$begingroup$
I have to conclude if the following problem is convergent or not using the root test or ratio test for positive series.
$$frac{(n!)^2}{(2n)!}$$
I've shown it to be convergent using the ratio test, but I've run into a problem when using the root test. I've concluded that $sqrt[n]{(n!)} to 1$ but Wolfram Alpha claims that $sqrt[n]{(n!)} to infty$.
Here is my work:
I started by reducing the expression:
$$frac{(n!)^2}{(2n)!} = frac{(n!)(n!)}{(2n)!}.$$
We have that
$$(n!) = (n)(n-1)(n-2)cdot...cdot 1 $$
$$(2n)! = (2n)(2n-1)cdot...cdot 1 $$
$$ = 2(n)(n-1/2)(n-1)(n-3/2)(n-2)cdot...cdot 1 $$ we rearrange this
$$ 2(n)(n-1)(n-2)(n-1/2)(n-3/2)cdot ...cdot 1 .$$ It should be clear that this "contains" $(n!).$ So now we have:
$$frac{(n!)(n!)}{(2n)!} = frac{(n!)}{(2n-1)(2n-3)cdot ...cdot 1 }. $$ I now applied the root test to this:
$$sqrt[n]{frac{(n!)}{(2n-1)(2-3)cdot ...cdot 1}} = {frac{sqrt[n]{(n!)}}{sqrt[n]{(2n-1)(2-3)cdot ...cdot 1}}}. $$
I examined $sqrt[n]{(n)}$:
$$sqrt[n]{(n)} = e^{(ln(n)cdot(1/n))}. $$ $e^x$ is continuous so we examine
$$lim_{nto infty}(frac{ln(n)}{n}) to frac{infty}{infty}. $$
We use L'hopital:
$$lim_{nto infty}(frac{ln(n)'}{n'}) = lim_{nto infty}(frac{1/n}{1}) = lim_{nto infty}1/n to 0.$$
So we have
$$lim_{nto infty}sqrt[n]{(n!)} = e^0 = 1.$$
We will have the same result given $sqrt[n]{(n-k)}.$
We can now return to $sqrt[n]{(n!)}$:
$$sqrt[n]{(n!)} = sqrt[n]{(n)}sqrt[n]{(n-1)}sqrt[n]{(n-2)}*..*1$$
$$lim_{nto infty}sqrt[n]{(n!)} = 1cdot1cdot1cdot ... cdot 1 = 1 .$$ I end up concluding the same thing regarding $$lim_{nto infty}sqrt[n]{{(2n-1)(2-3)cdot ... cdot 1}}.$$
I can't figure out why $sqrt[n]{(n!)} to infty,$ according to Wolfram Alpha. Is it because it becomes an indeterminant form of $1^infty$ ? Or have I simply made a mistake so that it never becomes $1cdot 1cdot 1cdot ... cdot 1,$ repeating infinitely?
I've looked at some of the other answers concerning the question of why $sqrt[n]{(n!)} to infty$ it doesn't answer why my specific argument is incorrect, and most of them seem to be using a "trick". Given what is in the book I'm using, these seem needlessly complicated or foreign to the reader. I am simply doing this for fun, the problem is already solved by using the ratio test, so perhaps I am simply underestimating the complexity of this problem.
real-analysis sequences-and-series limits
edited Jan 21 at 0:10
jordan_glen
1
asked Jan 20 at 22:22
DimajoDimajo
42
$endgroup$
2
$begingroup$
The line before "i end up concluding..." is incorrect: you can't apply arithmetic of limits to an expression in which the number of factors depends on $;n;$ itself! Only to an expression when the number of factors is a constant.
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– DonAntonio
Jan 20 at 22:31
1
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$lim_{nto infty} sqrt[n]{n} = 1$. Why would you think $lim_{nto infty} sqrt[n]{n!} = 1,$ too?
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– jordan_glen
Jan 20 at 22:32
$begingroup$
Also, please write well on this website, and not as you would writing a text message to a friend. That is, there is no word "i". There is a word "I". Also, the use of punctuation is immensely helpful, else your post comes across as one huge run-on sentence. Be respectful to other users using this site, particularly if you want to be understood, by taking care in what and how you write your questions.
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– jordan_glen
Jan 20 at 22:38
$begingroup$
The comment by @DonAntonio is the only one that answers what's actually wrong with the argument given in the question. All the answers given so far are basically repetition of what's already in this old question from 2012: math.stackexchange.com/questions/136626/…
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– Hans Lundmark
Jan 21 at 7:46
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@jordan_glen I took an extensive amount of time to try and write this as well as I could, I did attempt to capitilize i every time, but i guess i missed it? I was taught that you should never punctuate math, I don't see how it could ever be misunderstood without punctuation since there are clear breaks in the text or calculations. Why do you attribute this to malice and not incompetence? I tried my best and will follow your directions to the best of my ability in the future.
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– Dimajo
Jan 21 at 12:19
|
show 2 more comments
$begingroup$
I have to conclude if the following problem is convergent or not using the root test or ratio test for positive series.
$$frac{(n!)^2}{(2n)!}$$
I've shown it to be convergent using the ratio test, but I've run into a problem when using the root test. I've concluded that $sqrt[n]{(n!)} to 1$ but Wolfram Alpha claims that $sqrt[n]{(n!)} to infty$.
Here is my work:
I started by reducing the expression:
$$frac{(n!)^2}{(2n)!} = frac{(n!)(n!)}{(2n)!}.$$
We have that
$$(n!) = (n)(n-1)(n-2)cdot...cdot 1 $$
$$(2n)! = (2n)(2n-1)cdot...cdot 1 $$
$$ = 2(n)(n-1/2)(n-1)(n-3/2)(n-2)cdot...cdot 1 $$ we rearrange this
$$ 2(n)(n-1)(n-2)(n-1/2)(n-3/2)cdot ...cdot 1 .$$ It should be clear that this "contains" $(n!).$ So now we have:
$$frac{(n!)(n!)}{(2n)!} = frac{(n!)}{(2n-1)(2n-3)cdot ...cdot 1 }. $$ I now applied the root test to this:
$$sqrt[n]{frac{(n!)}{(2n-1)(2-3)cdot ...cdot 1}} = {frac{sqrt[n]{(n!)}}{sqrt[n]{(2n-1)(2-3)cdot ...cdot 1}}}. $$
I examined $sqrt[n]{(n)}$:
$$sqrt[n]{(n)} = e^{(ln(n)cdot(1/n))}. $$ $e^x$ is continuous so we examine
$$lim_{nto infty}(frac{ln(n)}{n}) to frac{infty}{infty}. $$
We use L'hopital:
$$lim_{nto infty}(frac{ln(n)'}{n'}) = lim_{nto infty}(frac{1/n}{1}) = lim_{nto infty}1/n to 0.$$
So we have
$$lim_{nto infty}sqrt[n]{(n!)} = e^0 = 1.$$
We will have the same result given $sqrt[n]{(n-k)}.$
We can now return to $sqrt[n]{(n!)}$:
$$sqrt[n]{(n!)} = sqrt[n]{(n)}sqrt[n]{(n-1)}sqrt[n]{(n-2)}*..*1$$
$$lim_{nto infty}sqrt[n]{(n!)} = 1cdot1cdot1cdot ... cdot 1 = 1 .$$ I end up concluding the same thing regarding $$lim_{nto infty}sqrt[n]{{(2n-1)(2-3)cdot ... cdot 1}}.$$
I can't figure out why $sqrt[n]{(n!)} to infty,$ according to Wolfram Alpha. Is it because it becomes an indeterminant form of $1^infty$ ? Or have I simply made a mistake so that it never becomes $1cdot 1cdot 1cdot ... cdot 1,$ repeating infinitely?
I've looked at some of the other answers concerning the question of why $sqrt[n]{(n!)} to infty$ it doesn't answer why my specific argument is incorrect, and most of them seem to be using a "trick". Given what is in the book I'm using, these seem needlessly complicated or foreign to the reader. I am simply doing this for fun, the problem is already solved by using the ratio test, so perhaps I am simply underestimating the complexity of this problem.
real-analysis sequences-and-series limits
edited Jan 21 at 0:10
jordan_glen
1
asked Jan 20 at 22:22
DimajoDimajo
42
$endgroup$
I have to conclude if the following problem is convergent or not using the root test or ratio test for positive series.
$$frac{(n!)^2}{(2n)!}$$
I've shown it to be convergent using the ratio test, but I've run into a problem when using the root test. I've concluded that $sqrt[n]{(n!)} to 1$ but Wolfram Alpha claims that $sqrt[n]{(n!)} to infty$.
Here is my work:
I started by reducing the expression:
$$frac{(n!)^2}{(2n)!} = frac{(n!)(n!)}{(2n)!}.$$
We have that
$$(n!) = (n)(n-1)(n-2)cdot...cdot 1 $$
$$(2n)! = (2n)(2n-1)cdot...cdot 1 $$
$$ = 2(n)(n-1/2)(n-1)(n-3/2)(n-2)cdot...cdot 1 $$ we rearrange this
$$ 2(n)(n-1)(n-2)(n-1/2)(n-3/2)cdot ...cdot 1 .$$ It should be clear that this "contains" $(n!).$ So now we have:
$$frac{(n!)(n!)}{(2n)!} = frac{(n!)}{(2n-1)(2n-3)cdot ...cdot 1 }. $$ I now applied the root test to this:
$$sqrt[n]{frac{(n!)}{(2n-1)(2-3)cdot ...cdot 1}} = {frac{sqrt[n]{(n!)}}{sqrt[n]{(2n-1)(2-3)cdot ...cdot 1}}}. $$
I examined $sqrt[n]{(n)}$:
$$sqrt[n]{(n)} = e^{(ln(n)cdot(1/n))}. $$ $e^x$ is continuous so we examine
$$lim_{nto infty}(frac{ln(n)}{n}) to frac{infty}{infty}. $$
We use L'hopital:
$$lim_{nto infty}(frac{ln(n)'}{n'}) = lim_{nto infty}(frac{1/n}{1}) = lim_{nto infty}1/n to 0.$$
So we have
$$lim_{nto infty}sqrt[n]{(n!)} = e^0 = 1.$$
We will have the same result given $sqrt[n]{(n-k)}.$
We can now return to $sqrt[n]{(n!)}$:
$$sqrt[n]{(n!)} = sqrt[n]{(n)}sqrt[n]{(n-1)}sqrt[n]{(n-2)}*..*1$$
$$lim_{nto infty}sqrt[n]{(n!)} = 1cdot1cdot1cdot ... cdot 1 = 1 .$$ I end up concluding the same thing regarding $$lim_{nto infty}sqrt[n]{{(2n-1)(2-3)cdot ... cdot 1}}.$$
I can't figure out why $sqrt[n]{(n!)} to infty,$ according to Wolfram Alpha. Is it because it becomes an indeterminant form of $1^infty$ ? Or have I simply made a mistake so that it never becomes $1cdot 1cdot 1cdot ... cdot 1,$ repeating infinitely?
I've looked at some of the other answers concerning the question of why $sqrt[n]{(n!)} to infty$ it doesn't answer why my specific argument is incorrect, and most of them seem to be using a "trick". Given what is in the book I'm using, these seem needlessly complicated or foreign to the reader. I am simply doing this for fun, the problem is already solved by using the ratio test, so perhaps I am simply underestimating the complexity of this problem.
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
edited Jan 21 at 0:10
jordan_glen
1
asked Jan 20 at 22:22
DimajoDimajo
42
edited Jan 21 at 0:10
jordan_glen
1
asked Jan 20 at 22:22
DimajoDimajo
42
edited Jan 21 at 0:10
jordan_glen
1
edited Jan 21 at 0:10
jordan_glen
1
edited Jan 21 at 0:10
jordan_glen
1
1
asked Jan 20 at 22:22
DimajoDimajo
42
asked Jan 20 at 22:22
DimajoDimajo
42
asked Jan 20 at 22:22
DimajoDimajo
42
42
2
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The line before "i end up concluding..." is incorrect: you can't apply arithmetic of limits to an expression in which the number of factors depends on $;n;$ itself! Only to an expression when the number of factors is a constant.
$endgroup$
– DonAntonio
Jan 20 at 22:31
1
$begingroup$
$lim_{nto infty} sqrt[n]{n} = 1$. Why would you think $lim_{nto infty} sqrt[n]{n!} = 1,$ too?
$endgroup$
– jordan_glen
Jan 20 at 22:32
$begingroup$
Also, please write well on this website, and not as you would writing a text message to a friend. That is, there is no word "i". There is a word "I". Also, the use of punctuation is immensely helpful, else your post comes across as one huge run-on sentence. Be respectful to other users using this site, particularly if you want to be understood, by taking care in what and how you write your questions.
$endgroup$
– jordan_glen
Jan 20 at 22:38
$begingroup$
The comment by @DonAntonio is the only one that answers what's actually wrong with the argument given in the question. All the answers given so far are basically repetition of what's already in this old question from 2012: math.stackexchange.com/questions/136626/…
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– Hans Lundmark
Jan 21 at 7:46
$begingroup$
@jordan_glen I took an extensive amount of time to try and write this as well as I could, I did attempt to capitilize i every time, but i guess i missed it? I was taught that you should never punctuate math, I don't see how it could ever be misunderstood without punctuation since there are clear breaks in the text or calculations. Why do you attribute this to malice and not incompetence? I tried my best and will follow your directions to the best of my ability in the future.
$endgroup$
– Dimajo
Jan 21 at 12:19
|
show 2 more comments
2
$begingroup$
The line before "i end up concluding..." is incorrect: you can't apply arithmetic of limits to an expression in which the number of factors depends on $;n;$ itself! Only to an expression when the number of factors is a constant.
$endgroup$
– DonAntonio
Jan 20 at 22:31
1
$begingroup$
$lim_{nto infty} sqrt[n]{n} = 1$. Why would you think $lim_{nto infty} sqrt[n]{n!} = 1,$ too?
$endgroup$
– jordan_glen
Jan 20 at 22:32
$begingroup$
Also, please write well on this website, and not as you would writing a text message to a friend. That is, there is no word "i". There is a word "I". Also, the use of punctuation is immensely helpful, else your post comes across as one huge run-on sentence. Be respectful to other users using this site, particularly if you want to be understood, by taking care in what and how you write your questions.
$endgroup$
– jordan_glen
Jan 20 at 22:38
$begingroup$
The comment by @DonAntonio is the only one that answers what's actually wrong with the argument given in the question. All the answers given so far are basically repetition of what's already in this old question from 2012: math.stackexchange.com/questions/136626/…
$endgroup$
– Hans Lundmark
Jan 21 at 7:46
$begingroup$
@jordan_glen I took an extensive amount of time to try and write this as well as I could, I did attempt to capitilize i every time, but i guess i missed it? I was taught that you should never punctuate math, I don't see how it could ever be misunderstood without punctuation since there are clear breaks in the text or calculations. Why do you attribute this to malice and not incompetence? I tried my best and will follow your directions to the best of my ability in the future.
$endgroup$
– Dimajo
Jan 21 at 12:19
2
2
$begingroup$
The line before "i end up concluding..." is incorrect: you can't apply arithmetic of limits to an expression in which the number of factors depends on $;n;$ itself! Only to an expression when the number of factors is a constant.
$endgroup$
– DonAntonio
Jan 20 at 22:31
$begingroup$
The line before "i end up concluding..." is incorrect: you can't apply arithmetic of limits to an expression in which the number of factors depends on $;n;$ itself! Only to an expression when the number of factors is a constant.
$endgroup$
– DonAntonio
Jan 20 at 22:31
1
1
$begingroup$
$lim_{nto infty} sqrt[n]{n} = 1$. Why would you think $lim_{nto infty} sqrt[n]{n!} = 1,$ too?
$endgroup$
– jordan_glen
Jan 20 at 22:32
$begingroup$
$lim_{nto infty} sqrt[n]{n} = 1$. Why would you think $lim_{nto infty} sqrt[n]{n!} = 1,$ too?
$endgroup$
– jordan_glen
Jan 20 at 22:32
$begingroup$
Also, please write well on this website, and not as you would writing a text message to a friend. That is, there is no word "i". There is a word "I". Also, the use of punctuation is immensely helpful, else your post comes across as one huge run-on sentence. Be respectful to other users using this site, particularly if you want to be understood, by taking care in what and how you write your questions.
$endgroup$
– jordan_glen
Jan 20 at 22:38
$begingroup$
Also, please write well on this website, and not as you would writing a text message to a friend. That is, there is no word "i". There is a word "I". Also, the use of punctuation is immensely helpful, else your post comes across as one huge run-on sentence. Be respectful to other users using this site, particularly if you want to be understood, by taking care in what and how you write your questions.
$endgroup$
– jordan_glen
Jan 20 at 22:38
$begingroup$
The comment by @DonAntonio is the only one that answers what's actually wrong with the argument given in the question. All the answers given so far are basically repetition of what's already in this old question from 2012: math.stackexchange.com/questions/136626/…
$endgroup$
– Hans Lundmark
Jan 21 at 7:46
$begingroup$
The comment by @DonAntonio is the only one that answers what's actually wrong with the argument given in the question. All the answers given so far are basically repetition of what's already in this old question from 2012: math.stackexchange.com/questions/136626/…
$endgroup$
– Hans Lundmark
Jan 21 at 7:46
$begingroup$
@jordan_glen I took an extensive amount of time to try and write this as well as I could, I did attempt to capitilize i every time, but i guess i missed it? I was taught that you should never punctuate math, I don't see how it could ever be misunderstood without punctuation since there are clear breaks in the text or calculations. Why do you attribute this to malice and not incompetence? I tried my best and will follow your directions to the best of my ability in the future.
$endgroup$
– Dimajo
Jan 21 at 12:19
$begingroup$
@jordan_glen I took an extensive amount of time to try and write this as well as I could, I did attempt to capitilize i every time, but i guess i missed it? I was taught that you should never punctuate math, I don't see how it could ever be misunderstood without punctuation since there are clear breaks in the text or calculations. Why do you attribute this to malice and not incompetence? I tried my best and will follow your directions to the best of my ability in the future.
$endgroup$
– Dimajo
Jan 21 at 12:19
|
show 2 more comments
6 Answers
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In the list $1,2,3,dots,n,$ the number of terms greater than $n/2$ is at least $n/2.$ Thus $n!>(n/2)^{n/2}.$ It follows that
$$(n!)^{1/n} > [(n/2)^{n/2}]^{1/n} = (n/2)^{1/2} to infty.$$
answered Jan 21 at 0:25
zhw.zhw.
73.8k43175
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add a comment |
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Let $a_{n}=sqrt[n]{n!}$ and $b_{n}=ln a_{n}$. Then $b_{n}=frac{1}{n}sum_{k=1}^{n}ln k$.
For $kgeq2$ and $xin[k-1,k]$, we have $ln kgeqln x$, so $ln k=int_{k-1}^{k}ln k,dxgeqint_{k-1}^{k}ln x,dx$.
Therefore
begin{eqnarray*}
b_{n} & geq & frac{1}{n}sum_{k=2}^{n}ln k\
& geq & frac{1}{n}sum_{k=2}^{n}int_{k-1}^{k}ln x,dx\
& = & frac{1}{n}int_{1}^{n}ln x,dx\
& = & frac{1}{n}left(nln n-n-1right)\
& = & ln n-frac{n+1}{n}\
& rightarrow & infty
end{eqnarray*}
as $nrightarrowinfty$. Therefore $a_{n}=exp(b_{n})rightarrowinfty$
as $nrightarrowinfty$.
answered Jan 20 at 22:38
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,40138
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Per Stirling's formula that states
$$ n! sim {n^n e^{-n}over sqrt{2pi n}},$$
you have
$$root{n}of{n!} sim left({n^n e^{-n}sqrt{2pi n}}right)^{1/n}
sim {nover e} $$
as $ntoinfty$.
answered Jan 20 at 22:48
ncmathsadistncmathsadist
42.9k260103
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2
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The solution is not self-contained because what has been assumed (i.e., Stirling's formula) is far more complicated and involving than what has been asked to prove (i.e. $sqrt[n]{n!}rightarrowinfty$).
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– Danny Pak-Keung Chan
Jan 20 at 22:51
1
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If OP knew Stirling's formula, he would be mathematically mature enough and should not get stuck in that simple question.
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– Danny Pak-Keung Chan
Jan 20 at 23:00
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The $sqrt{2pi n}$ belongs in the numerator, not the denominator.
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– Ross Millikan
Jan 21 at 0:12
add a comment |
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You may consider the reciprocals $boxed{frac{1}{sqrt[n]{n!}}}$ and proceed as follows using GM-HM (inequality between geometric and harmonic mean) and using
$sum_{k=1}^nfrac{1}{k} < ln n +1$ for $n >1$
- $frac{ln n}{n } stackrel{n to infty}{longrightarrow} 0$
begin{eqnarray*} frac{1}{sqrt[n]{n!} }
& stackrel{GM-HM}{leq} & frac{sum_{k=1}^nfrac{1}{k}}{n}\
& stackrel{sum_{k=1}^nfrac{1}{k}< ln n + 1}{leq} & frac{ln n +1}{n}\
& stackrel{n to infty}{longrightarrow} & 0\
end{eqnarray*}
It follows $boxed{sqrt[n]{n!} stackrel{n to infty}{longrightarrow} infty}$
answered Jan 21 at 4:25
trancelocationtrancelocation
12.6k1826
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add a comment |
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Start with $$a_n=sqrt[n]{n!} implies log(a_n)=frac 1n log(n!)$$ Now, by Stirling approximation
$$log(n!)=n (log (n)-1)+frac{1}{2} left(log (2 pi )+log
left({n}right)right)+frac{1}{12
n}+Oleft(frac{1}{n^3}right)$$
$$log(a_n)=log (n)-1+frac{1}{2n} left(log (2 pi )+log
left({n}right)right)+Oleft(frac{1}{n^2}right)$$ So, by the end
$$sqrt[n]{n!}simeq frac n e$$
answered Jan 21 at 3:58
Claude LeiboviciClaude Leibovici
123k1157135
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This is not rigorous, but it gives an intuitive solution (and this can be made rigorous). From Calculus we know that:
$$
log(n!) = sum_{k=1}^{n} ln(k) approx int_1^n ln(x) dx = n ln n - n + C.
$$
Therefore,
$$
n! = e^{log(n!)} approx e^{n ln n - n + C} approx e^{n ln n}e^{-n} = left( frac{n}{e}right)^n.
$$
answered Jan 21 at 4:12
JavaManJavaMan
11.1k12755
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I should note that this essentially the same as Danny Pak-Keung Chen's solution (or at least it's the same idea), though his solution is more rigorous and thus less intuitive.
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– JavaMan
Jan 21 at 4:19
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$begingroup$
In the list $1,2,3,dots,n,$ the number of terms greater than $n/2$ is at least $n/2.$ Thus $n!>(n/2)^{n/2}.$ It follows that
$$(n!)^{1/n} > [(n/2)^{n/2}]^{1/n} = (n/2)^{1/2} to infty.$$
answered Jan 21 at 0:25
zhw.zhw.
73.8k43175
$endgroup$
add a comment |
$begingroup$
In the list $1,2,3,dots,n,$ the number of terms greater than $n/2$ is at least $n/2.$ Thus $n!>(n/2)^{n/2}.$ It follows that
$$(n!)^{1/n} > [(n/2)^{n/2}]^{1/n} = (n/2)^{1/2} to infty.$$
answered Jan 21 at 0:25
zhw.zhw.
73.8k43175
$endgroup$
add a comment |
$begingroup$
In the list $1,2,3,dots,n,$ the number of terms greater than $n/2$ is at least $n/2.$ Thus $n!>(n/2)^{n/2}.$ It follows that
$$(n!)^{1/n} > [(n/2)^{n/2}]^{1/n} = (n/2)^{1/2} to infty.$$
answered Jan 21 at 0:25
zhw.zhw.
73.8k43175
$endgroup$
In the list $1,2,3,dots,n,$ the number of terms greater than $n/2$ is at least $n/2.$ Thus $n!>(n/2)^{n/2}.$ It follows that
$$(n!)^{1/n} > [(n/2)^{n/2}]^{1/n} = (n/2)^{1/2} to infty.$$
answered Jan 21 at 0:25
zhw.zhw.
73.8k43175
answered Jan 21 at 0:25
zhw.zhw.
73.8k43175
answered Jan 21 at 0:25
zhw.zhw.
73.8k43175
answered Jan 21 at 0:25
zhw.zhw.
73.8k43175
73.8k43175
add a comment |
add a comment |
$begingroup$
Let $a_{n}=sqrt[n]{n!}$ and $b_{n}=ln a_{n}$. Then $b_{n}=frac{1}{n}sum_{k=1}^{n}ln k$.
For $kgeq2$ and $xin[k-1,k]$, we have $ln kgeqln x$, so $ln k=int_{k-1}^{k}ln k,dxgeqint_{k-1}^{k}ln x,dx$.
Therefore
begin{eqnarray*}
b_{n} & geq & frac{1}{n}sum_{k=2}^{n}ln k\
& geq & frac{1}{n}sum_{k=2}^{n}int_{k-1}^{k}ln x,dx\
& = & frac{1}{n}int_{1}^{n}ln x,dx\
& = & frac{1}{n}left(nln n-n-1right)\
& = & ln n-frac{n+1}{n}\
& rightarrow & infty
end{eqnarray*}
as $nrightarrowinfty$. Therefore $a_{n}=exp(b_{n})rightarrowinfty$
as $nrightarrowinfty$.
answered Jan 20 at 22:38
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,40138
$endgroup$
add a comment |
$begingroup$
Let $a_{n}=sqrt[n]{n!}$ and $b_{n}=ln a_{n}$. Then $b_{n}=frac{1}{n}sum_{k=1}^{n}ln k$.
For $kgeq2$ and $xin[k-1,k]$, we have $ln kgeqln x$, so $ln k=int_{k-1}^{k}ln k,dxgeqint_{k-1}^{k}ln x,dx$.
Therefore
begin{eqnarray*}
b_{n} & geq & frac{1}{n}sum_{k=2}^{n}ln k\
& geq & frac{1}{n}sum_{k=2}^{n}int_{k-1}^{k}ln x,dx\
& = & frac{1}{n}int_{1}^{n}ln x,dx\
& = & frac{1}{n}left(nln n-n-1right)\
& = & ln n-frac{n+1}{n}\
& rightarrow & infty
end{eqnarray*}
as $nrightarrowinfty$. Therefore $a_{n}=exp(b_{n})rightarrowinfty$
as $nrightarrowinfty$.
answered Jan 20 at 22:38
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,40138
$endgroup$
add a comment |
$begingroup$
Let $a_{n}=sqrt[n]{n!}$ and $b_{n}=ln a_{n}$. Then $b_{n}=frac{1}{n}sum_{k=1}^{n}ln k$.
For $kgeq2$ and $xin[k-1,k]$, we have $ln kgeqln x$, so $ln k=int_{k-1}^{k}ln k,dxgeqint_{k-1}^{k}ln x,dx$.
Therefore
begin{eqnarray*}
b_{n} & geq & frac{1}{n}sum_{k=2}^{n}ln k\
& geq & frac{1}{n}sum_{k=2}^{n}int_{k-1}^{k}ln x,dx\
& = & frac{1}{n}int_{1}^{n}ln x,dx\
& = & frac{1}{n}left(nln n-n-1right)\
& = & ln n-frac{n+1}{n}\
& rightarrow & infty
end{eqnarray*}
as $nrightarrowinfty$. Therefore $a_{n}=exp(b_{n})rightarrowinfty$
as $nrightarrowinfty$.
answered Jan 20 at 22:38
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,40138
$endgroup$
Let $a_{n}=sqrt[n]{n!}$ and $b_{n}=ln a_{n}$. Then $b_{n}=frac{1}{n}sum_{k=1}^{n}ln k$.
For $kgeq2$ and $xin[k-1,k]$, we have $ln kgeqln x$, so $ln k=int_{k-1}^{k}ln k,dxgeqint_{k-1}^{k}ln x,dx$.
Therefore
begin{eqnarray*}
b_{n} & geq & frac{1}{n}sum_{k=2}^{n}ln k\
& geq & frac{1}{n}sum_{k=2}^{n}int_{k-1}^{k}ln x,dx\
& = & frac{1}{n}int_{1}^{n}ln x,dx\
& = & frac{1}{n}left(nln n-n-1right)\
& = & ln n-frac{n+1}{n}\
& rightarrow & infty
end{eqnarray*}
as $nrightarrowinfty$. Therefore $a_{n}=exp(b_{n})rightarrowinfty$
as $nrightarrowinfty$.
answered Jan 20 at 22:38
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,40138
answered Jan 20 at 22:38
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,40138
answered Jan 20 at 22:38
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,40138
answered Jan 20 at 22:38
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,40138
2,40138
add a comment |
add a comment |
$begingroup$
Per Stirling's formula that states
$$ n! sim {n^n e^{-n}over sqrt{2pi n}},$$
you have
$$root{n}of{n!} sim left({n^n e^{-n}sqrt{2pi n}}right)^{1/n}
sim {nover e} $$
as $ntoinfty$.
answered Jan 20 at 22:48
ncmathsadistncmathsadist
42.9k260103
$endgroup$
2
$begingroup$
The solution is not self-contained because what has been assumed (i.e., Stirling's formula) is far more complicated and involving than what has been asked to prove (i.e. $sqrt[n]{n!}rightarrowinfty$).
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 22:51
1
$begingroup$
If OP knew Stirling's formula, he would be mathematically mature enough and should not get stuck in that simple question.
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 23:00
$begingroup$
The $sqrt{2pi n}$ belongs in the numerator, not the denominator.
$endgroup$
– Ross Millikan
Jan 21 at 0:12
add a comment |
$begingroup$
Per Stirling's formula that states
$$ n! sim {n^n e^{-n}over sqrt{2pi n}},$$
you have
$$root{n}of{n!} sim left({n^n e^{-n}sqrt{2pi n}}right)^{1/n}
sim {nover e} $$
as $ntoinfty$.
answered Jan 20 at 22:48
ncmathsadistncmathsadist
42.9k260103
$endgroup$
2
$begingroup$
The solution is not self-contained because what has been assumed (i.e., Stirling's formula) is far more complicated and involving than what has been asked to prove (i.e. $sqrt[n]{n!}rightarrowinfty$).
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 22:51
1
$begingroup$
If OP knew Stirling's formula, he would be mathematically mature enough and should not get stuck in that simple question.
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 23:00
$begingroup$
The $sqrt{2pi n}$ belongs in the numerator, not the denominator.
$endgroup$
– Ross Millikan
Jan 21 at 0:12
add a comment |
$begingroup$
Per Stirling's formula that states
$$ n! sim {n^n e^{-n}over sqrt{2pi n}},$$
you have
$$root{n}of{n!} sim left({n^n e^{-n}sqrt{2pi n}}right)^{1/n}
sim {nover e} $$
as $ntoinfty$.
answered Jan 20 at 22:48
ncmathsadistncmathsadist
42.9k260103
$endgroup$
Per Stirling's formula that states
$$ n! sim {n^n e^{-n}over sqrt{2pi n}},$$
you have
$$root{n}of{n!} sim left({n^n e^{-n}sqrt{2pi n}}right)^{1/n}
sim {nover e} $$
as $ntoinfty$.
answered Jan 20 at 22:48
ncmathsadistncmathsadist
42.9k260103
edited Jan 21 at 0:52
answered Jan 20 at 22:48
ncmathsadistncmathsadist
42.9k260103
answered Jan 20 at 22:48
ncmathsadistncmathsadist
42.9k260103
answered Jan 20 at 22:48
ncmathsadistncmathsadist
42.9k260103
42.9k260103
2
$begingroup$
The solution is not self-contained because what has been assumed (i.e., Stirling's formula) is far more complicated and involving than what has been asked to prove (i.e. $sqrt[n]{n!}rightarrowinfty$).
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 22:51
1
$begingroup$
If OP knew Stirling's formula, he would be mathematically mature enough and should not get stuck in that simple question.
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 23:00
$begingroup$
The $sqrt{2pi n}$ belongs in the numerator, not the denominator.
$endgroup$
– Ross Millikan
Jan 21 at 0:12
add a comment |
2
$begingroup$
The solution is not self-contained because what has been assumed (i.e., Stirling's formula) is far more complicated and involving than what has been asked to prove (i.e. $sqrt[n]{n!}rightarrowinfty$).
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 22:51
1
$begingroup$
If OP knew Stirling's formula, he would be mathematically mature enough and should not get stuck in that simple question.
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 23:00
$begingroup$
The $sqrt{2pi n}$ belongs in the numerator, not the denominator.
$endgroup$
– Ross Millikan
Jan 21 at 0:12
2
2
$begingroup$
The solution is not self-contained because what has been assumed (i.e., Stirling's formula) is far more complicated and involving than what has been asked to prove (i.e. $sqrt[n]{n!}rightarrowinfty$).
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 22:51
$begingroup$
The solution is not self-contained because what has been assumed (i.e., Stirling's formula) is far more complicated and involving than what has been asked to prove (i.e. $sqrt[n]{n!}rightarrowinfty$).
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 22:51
1
1
$begingroup$
If OP knew Stirling's formula, he would be mathematically mature enough and should not get stuck in that simple question.
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 23:00
$begingroup$
If OP knew Stirling's formula, he would be mathematically mature enough and should not get stuck in that simple question.
$endgroup$
– Danny Pak-Keung Chan
Jan 20 at 23:00
$begingroup$
The $sqrt{2pi n}$ belongs in the numerator, not the denominator.
$endgroup$
– Ross Millikan
Jan 21 at 0:12
$begingroup$
The $sqrt{2pi n}$ belongs in the numerator, not the denominator.
$endgroup$
– Ross Millikan
Jan 21 at 0:12
add a comment |
$begingroup$
You may consider the reciprocals $boxed{frac{1}{sqrt[n]{n!}}}$ and proceed as follows using GM-HM (inequality between geometric and harmonic mean) and using
$sum_{k=1}^nfrac{1}{k} < ln n +1$ for $n >1$
- $frac{ln n}{n } stackrel{n to infty}{longrightarrow} 0$
begin{eqnarray*} frac{1}{sqrt[n]{n!} }
& stackrel{GM-HM}{leq} & frac{sum_{k=1}^nfrac{1}{k}}{n}\
& stackrel{sum_{k=1}^nfrac{1}{k}< ln n + 1}{leq} & frac{ln n +1}{n}\
& stackrel{n to infty}{longrightarrow} & 0\
end{eqnarray*}
It follows $boxed{sqrt[n]{n!} stackrel{n to infty}{longrightarrow} infty}$
answered Jan 21 at 4:25
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
You may consider the reciprocals $boxed{frac{1}{sqrt[n]{n!}}}$ and proceed as follows using GM-HM (inequality between geometric and harmonic mean) and using
$sum_{k=1}^nfrac{1}{k} < ln n +1$ for $n >1$
- $frac{ln n}{n } stackrel{n to infty}{longrightarrow} 0$
begin{eqnarray*} frac{1}{sqrt[n]{n!} }
& stackrel{GM-HM}{leq} & frac{sum_{k=1}^nfrac{1}{k}}{n}\
& stackrel{sum_{k=1}^nfrac{1}{k}< ln n + 1}{leq} & frac{ln n +1}{n}\
& stackrel{n to infty}{longrightarrow} & 0\
end{eqnarray*}
It follows $boxed{sqrt[n]{n!} stackrel{n to infty}{longrightarrow} infty}$
answered Jan 21 at 4:25
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
You may consider the reciprocals $boxed{frac{1}{sqrt[n]{n!}}}$ and proceed as follows using GM-HM (inequality between geometric and harmonic mean) and using
$sum_{k=1}^nfrac{1}{k} < ln n +1$ for $n >1$
- $frac{ln n}{n } stackrel{n to infty}{longrightarrow} 0$
begin{eqnarray*} frac{1}{sqrt[n]{n!} }
& stackrel{GM-HM}{leq} & frac{sum_{k=1}^nfrac{1}{k}}{n}\
& stackrel{sum_{k=1}^nfrac{1}{k}< ln n + 1}{leq} & frac{ln n +1}{n}\
& stackrel{n to infty}{longrightarrow} & 0\
end{eqnarray*}
It follows $boxed{sqrt[n]{n!} stackrel{n to infty}{longrightarrow} infty}$
answered Jan 21 at 4:25
trancelocationtrancelocation
12.6k1826
$endgroup$
You may consider the reciprocals $boxed{frac{1}{sqrt[n]{n!}}}$ and proceed as follows using GM-HM (inequality between geometric and harmonic mean) and using
$sum_{k=1}^nfrac{1}{k} < ln n +1$ for $n >1$
- $frac{ln n}{n } stackrel{n to infty}{longrightarrow} 0$
begin{eqnarray*} frac{1}{sqrt[n]{n!} }
& stackrel{GM-HM}{leq} & frac{sum_{k=1}^nfrac{1}{k}}{n}\
& stackrel{sum_{k=1}^nfrac{1}{k}< ln n + 1}{leq} & frac{ln n +1}{n}\
& stackrel{n to infty}{longrightarrow} & 0\
end{eqnarray*}
It follows $boxed{sqrt[n]{n!} stackrel{n to infty}{longrightarrow} infty}$
answered Jan 21 at 4:25
trancelocationtrancelocation
12.6k1826
answered Jan 21 at 4:25
trancelocationtrancelocation
12.6k1826
answered Jan 21 at 4:25
trancelocationtrancelocation
12.6k1826
answered Jan 21 at 4:25
trancelocationtrancelocation
12.6k1826
12.6k1826
add a comment |
add a comment |
$begingroup$
Start with $$a_n=sqrt[n]{n!} implies log(a_n)=frac 1n log(n!)$$ Now, by Stirling approximation
$$log(n!)=n (log (n)-1)+frac{1}{2} left(log (2 pi )+log
left({n}right)right)+frac{1}{12
n}+Oleft(frac{1}{n^3}right)$$
$$log(a_n)=log (n)-1+frac{1}{2n} left(log (2 pi )+log
left({n}right)right)+Oleft(frac{1}{n^2}right)$$ So, by the end
$$sqrt[n]{n!}simeq frac n e$$
answered Jan 21 at 3:58
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
add a comment |
$begingroup$
Start with $$a_n=sqrt[n]{n!} implies log(a_n)=frac 1n log(n!)$$ Now, by Stirling approximation
$$log(n!)=n (log (n)-1)+frac{1}{2} left(log (2 pi )+log
left({n}right)right)+frac{1}{12
n}+Oleft(frac{1}{n^3}right)$$
$$log(a_n)=log (n)-1+frac{1}{2n} left(log (2 pi )+log
left({n}right)right)+Oleft(frac{1}{n^2}right)$$ So, by the end
$$sqrt[n]{n!}simeq frac n e$$
answered Jan 21 at 3:58
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
add a comment |
$begingroup$
Start with $$a_n=sqrt[n]{n!} implies log(a_n)=frac 1n log(n!)$$ Now, by Stirling approximation
$$log(n!)=n (log (n)-1)+frac{1}{2} left(log (2 pi )+log
left({n}right)right)+frac{1}{12
n}+Oleft(frac{1}{n^3}right)$$
$$log(a_n)=log (n)-1+frac{1}{2n} left(log (2 pi )+log
left({n}right)right)+Oleft(frac{1}{n^2}right)$$ So, by the end
$$sqrt[n]{n!}simeq frac n e$$
answered Jan 21 at 3:58
Claude LeiboviciClaude Leibovici
123k1157135
$endgroup$
Start with $$a_n=sqrt[n]{n!} implies log(a_n)=frac 1n log(n!)$$ Now, by Stirling approximation
$$log(n!)=n (log (n)-1)+frac{1}{2} left(log (2 pi )+log
left({n}right)right)+frac{1}{12
n}+Oleft(frac{1}{n^3}right)$$
$$log(a_n)=log (n)-1+frac{1}{2n} left(log (2 pi )+log
left({n}right)right)+Oleft(frac{1}{n^2}right)$$ So, by the end
$$sqrt[n]{n!}simeq frac n e$$
answered Jan 21 at 3:58
Claude LeiboviciClaude Leibovici
123k1157135
answered Jan 21 at 3:58
Claude LeiboviciClaude Leibovici
123k1157135
answered Jan 21 at 3:58
Claude LeiboviciClaude Leibovici
123k1157135
answered Jan 21 at 3:58
Claude LeiboviciClaude Leibovici
123k1157135
123k1157135
add a comment |
add a comment |
$begingroup$
This is not rigorous, but it gives an intuitive solution (and this can be made rigorous). From Calculus we know that:
$$
log(n!) = sum_{k=1}^{n} ln(k) approx int_1^n ln(x) dx = n ln n - n + C.
$$
Therefore,
$$
n! = e^{log(n!)} approx e^{n ln n - n + C} approx e^{n ln n}e^{-n} = left( frac{n}{e}right)^n.
$$
answered Jan 21 at 4:12
JavaManJavaMan
11.1k12755
$endgroup$
$begingroup$
I should note that this essentially the same as Danny Pak-Keung Chen's solution (or at least it's the same idea), though his solution is more rigorous and thus less intuitive.
$endgroup$
– JavaMan
Jan 21 at 4:19
add a comment |
$begingroup$
This is not rigorous, but it gives an intuitive solution (and this can be made rigorous). From Calculus we know that:
$$
log(n!) = sum_{k=1}^{n} ln(k) approx int_1^n ln(x) dx = n ln n - n + C.
$$
Therefore,
$$
n! = e^{log(n!)} approx e^{n ln n - n + C} approx e^{n ln n}e^{-n} = left( frac{n}{e}right)^n.
$$
answered Jan 21 at 4:12
JavaManJavaMan
11.1k12755
$endgroup$
$begingroup$
I should note that this essentially the same as Danny Pak-Keung Chen's solution (or at least it's the same idea), though his solution is more rigorous and thus less intuitive.
$endgroup$
– JavaMan
Jan 21 at 4:19
add a comment |
$begingroup$
This is not rigorous, but it gives an intuitive solution (and this can be made rigorous). From Calculus we know that:
$$
log(n!) = sum_{k=1}^{n} ln(k) approx int_1^n ln(x) dx = n ln n - n + C.
$$
Therefore,
$$
n! = e^{log(n!)} approx e^{n ln n - n + C} approx e^{n ln n}e^{-n} = left( frac{n}{e}right)^n.
$$
answered Jan 21 at 4:12
JavaManJavaMan
11.1k12755
$endgroup$
This is not rigorous, but it gives an intuitive solution (and this can be made rigorous). From Calculus we know that:
$$
log(n!) = sum_{k=1}^{n} ln(k) approx int_1^n ln(x) dx = n ln n - n + C.
$$
Therefore,
$$
n! = e^{log(n!)} approx e^{n ln n - n + C} approx e^{n ln n}e^{-n} = left( frac{n}{e}right)^n.
$$
answered Jan 21 at 4:12
JavaManJavaMan
11.1k12755
answered Jan 21 at 4:12
JavaManJavaMan
11.1k12755
answered Jan 21 at 4:12
JavaManJavaMan
11.1k12755
answered Jan 21 at 4:12
JavaManJavaMan
11.1k12755
11.1k12755
$begingroup$
I should note that this essentially the same as Danny Pak-Keung Chen's solution (or at least it's the same idea), though his solution is more rigorous and thus less intuitive.
$endgroup$
– JavaMan
Jan 21 at 4:19
add a comment |
$begingroup$
I should note that this essentially the same as Danny Pak-Keung Chen's solution (or at least it's the same idea), though his solution is more rigorous and thus less intuitive.
$endgroup$
– JavaMan
Jan 21 at 4:19
$begingroup$
I should note that this essentially the same as Danny Pak-Keung Chen's solution (or at least it's the same idea), though his solution is more rigorous and thus less intuitive.
$endgroup$
– JavaMan
Jan 21 at 4:19
$begingroup$
I should note that this essentially the same as Danny Pak-Keung Chen's solution (or at least it's the same idea), though his solution is more rigorous and thus less intuitive.
$endgroup$
– JavaMan
Jan 21 at 4:19
add a comment |
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The line before "i end up concluding..." is incorrect: you can't apply arithmetic of limits to an expression in which the number of factors depends on $;n;$ itself! Only to an expression when the number of factors is a constant.
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– DonAntonio
Jan 20 at 22:31
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$lim_{nto infty} sqrt[n]{n} = 1$. Why would you think $lim_{nto infty} sqrt[n]{n!} = 1,$ too?
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– jordan_glen
Jan 20 at 22:32
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Also, please write well on this website, and not as you would writing a text message to a friend. That is, there is no word "i". There is a word "I". Also, the use of punctuation is immensely helpful, else your post comes across as one huge run-on sentence. Be respectful to other users using this site, particularly if you want to be understood, by taking care in what and how you write your questions.
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– jordan_glen
Jan 20 at 22:38
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The comment by @DonAntonio is the only one that answers what's actually wrong with the argument given in the question. All the answers given so far are basically repetition of what's already in this old question from 2012: math.stackexchange.com/questions/136626/…
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– Hans Lundmark
Jan 21 at 7:46
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@jordan_glen I took an extensive amount of time to try and write this as well as I could, I did attempt to capitilize i every time, but i guess i missed it? I was taught that you should never punctuate math, I don't see how it could ever be misunderstood without punctuation since there are clear breaks in the text or calculations. Why do you attribute this to malice and not incompetence? I tried my best and will follow your directions to the best of my ability in the future.
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– Dimajo
Jan 21 at 12:19