Find $lim_{nto infty}((n+1)!)^{frac{1}{n+1}}-((n)!)^{frac{1}{n}}.$ [duplicate]
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This question already has an answer here:
Limit of the sequence $a_n=sqrt[n+1]{(n+1)!}-sqrt[n]{n!}$
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Find $lim_{nto
infty}((n+1)!)^{frac{1}{n+1}}-((n)!)^{frac{1}{n}}.$
We need to deal the limit $lim_{nto infty} frac{log(1)+log(2)+...+log(n)}{n}$. We know that $lim_{nto infty} log(n)=infty implies lim_{nto infty} frac{log(1)+log(2)+...+log(n)}{n}=infty$(since, By Cauchy's first theorem on limit). Hence we get $infty-infty$. How do I show that there exists finite limit?
real-analysis sequences-and-series limits
asked Dec 6 '18 at 15:57
Unknown xUnknown x
2,51211026
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marked as duplicate by Paramanand Singh limits
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Dec 6 '18 at 16:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Limit of the sequence $a_n=sqrt[n+1]{(n+1)!}-sqrt[n]{n!}$
1 answer
Find $lim_{nto
infty}((n+1)!)^{frac{1}{n+1}}-((n)!)^{frac{1}{n}}.$
We need to deal the limit $lim_{nto infty} frac{log(1)+log(2)+...+log(n)}{n}$. We know that $lim_{nto infty} log(n)=infty implies lim_{nto infty} frac{log(1)+log(2)+...+log(n)}{n}=infty$(since, By Cauchy's first theorem on limit). Hence we get $infty-infty$. How do I show that there exists finite limit?
real-analysis sequences-and-series limits
asked Dec 6 '18 at 15:57
Unknown xUnknown x
2,51211026
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marked as duplicate by Paramanand Singh limits
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Dec 6 '18 at 16:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Let me check. But the limit is infinity.
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– Unknown x
Dec 6 '18 at 16:01
3
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The limit should be $1/e$.
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– Paramanand Singh
Dec 6 '18 at 16:04
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How do we get?Can you give some hint?
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– Unknown x
Dec 6 '18 at 16:05
add a comment |
$begingroup$
This question already has an answer here:
Limit of the sequence $a_n=sqrt[n+1]{(n+1)!}-sqrt[n]{n!}$
1 answer
Find $lim_{nto
infty}((n+1)!)^{frac{1}{n+1}}-((n)!)^{frac{1}{n}}.$
We need to deal the limit $lim_{nto infty} frac{log(1)+log(2)+...+log(n)}{n}$. We know that $lim_{nto infty} log(n)=infty implies lim_{nto infty} frac{log(1)+log(2)+...+log(n)}{n}=infty$(since, By Cauchy's first theorem on limit). Hence we get $infty-infty$. How do I show that there exists finite limit?
real-analysis sequences-and-series limits
asked Dec 6 '18 at 15:57
Unknown xUnknown x
2,51211026
$endgroup$
This question already has an answer here:
Limit of the sequence $a_n=sqrt[n+1]{(n+1)!}-sqrt[n]{n!}$
1 answer
Find $lim_{nto
infty}((n+1)!)^{frac{1}{n+1}}-((n)!)^{frac{1}{n}}.$
We need to deal the limit $lim_{nto infty} frac{log(1)+log(2)+...+log(n)}{n}$. We know that $lim_{nto infty} log(n)=infty implies lim_{nto infty} frac{log(1)+log(2)+...+log(n)}{n}=infty$(since, By Cauchy's first theorem on limit). Hence we get $infty-infty$. How do I show that there exists finite limit?
This question already has an answer here:
Limit of the sequence $a_n=sqrt[n+1]{(n+1)!}-sqrt[n]{n!}$
1 answer
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
asked Dec 6 '18 at 15:57
Unknown xUnknown x
2,51211026
asked Dec 6 '18 at 15:57
Unknown xUnknown x
2,51211026
edited Dec 6 '18 at 15:59
Unknown x
asked Dec 6 '18 at 15:57
Unknown xUnknown x
2,51211026
asked Dec 6 '18 at 15:57
Unknown xUnknown x
2,51211026
asked Dec 6 '18 at 15:57
Unknown xUnknown x
2,51211026
2,51211026
marked as duplicate by Paramanand Singh limits
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Dec 6 '18 at 16:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Paramanand Singh limits
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Dec 6 '18 at 16:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Let me check. But the limit is infinity.
$endgroup$
– Unknown x
Dec 6 '18 at 16:01
3
$begingroup$
The limit should be $1/e$.
$endgroup$
– Paramanand Singh
Dec 6 '18 at 16:04
$begingroup$
How do we get?Can you give some hint?
$endgroup$
– Unknown x
Dec 6 '18 at 16:05
add a comment |
$begingroup$
Let me check. But the limit is infinity.
$endgroup$
– Unknown x
Dec 6 '18 at 16:01
3
$begingroup$
The limit should be $1/e$.
$endgroup$
– Paramanand Singh
Dec 6 '18 at 16:04
$begingroup$
How do we get?Can you give some hint?
$endgroup$
– Unknown x
Dec 6 '18 at 16:05
$begingroup$
Let me check. But the limit is infinity.
$endgroup$
– Unknown x
Dec 6 '18 at 16:01
$begingroup$
Let me check. But the limit is infinity.
$endgroup$
– Unknown x
Dec 6 '18 at 16:01
3
3
$begingroup$
The limit should be $1/e$.
$endgroup$
– Paramanand Singh
Dec 6 '18 at 16:04
$begingroup$
The limit should be $1/e$.
$endgroup$
– Paramanand Singh
Dec 6 '18 at 16:04
$begingroup$
How do we get?Can you give some hint?
$endgroup$
– Unknown x
Dec 6 '18 at 16:05
$begingroup$
How do we get?Can you give some hint?
$endgroup$
– Unknown x
Dec 6 '18 at 16:05
add a comment |
1 Answer
1
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oldest
votes
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HINT:
Using Stirling's Formula we have
$$begin{align}
left((n+1)!right)^{1/(n+1)}-left(n!right)^{1/n}&=left(left(sqrt{2pi(n+1)}left(frac{n+1}{e}right)^{n+1}right)left(1+O(1/n)right)right)^{1/(n+1)}\\
&-left(left(sqrt{2pi n}left(frac{n}{e}right)^{n}right)left(1+O(1/n)right)right)^{1/n}
end{align}$$
Can you finish now?
Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
answered Dec 6 '18 at 16:05
Mark ViolaMark Viola
132k1275173
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$begingroup$
This it is not convincing at all. Did you really check that this gets somewhere ?
$endgroup$
– Ewan Delanoy
Dec 6 '18 at 16:13
1
$begingroup$
@EwanDelanoy First of all, it is a HINT only. And yes, I did check this and found that the limit is simply $1/e$.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:15
$begingroup$
@EwanDelanoy Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:22
$begingroup$
Is your sirling formula correct? $sqrt{2 pi n}$ in the numerator?
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– Unknown x
Dec 6 '18 at 17:23
$begingroup$
your alternate approach is very excellent :)
$endgroup$
– Unknown x
Dec 6 '18 at 17:25
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT:
Using Stirling's Formula we have
$$begin{align}
left((n+1)!right)^{1/(n+1)}-left(n!right)^{1/n}&=left(left(sqrt{2pi(n+1)}left(frac{n+1}{e}right)^{n+1}right)left(1+O(1/n)right)right)^{1/(n+1)}\\
&-left(left(sqrt{2pi n}left(frac{n}{e}right)^{n}right)left(1+O(1/n)right)right)^{1/n}
end{align}$$
Can you finish now?
Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
answered Dec 6 '18 at 16:05
Mark ViolaMark Viola
132k1275173
$endgroup$
$begingroup$
This it is not convincing at all. Did you really check that this gets somewhere ?
$endgroup$
– Ewan Delanoy
Dec 6 '18 at 16:13
1
$begingroup$
@EwanDelanoy First of all, it is a HINT only. And yes, I did check this and found that the limit is simply $1/e$.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:15
$begingroup$
@EwanDelanoy Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:22
$begingroup$
Is your sirling formula correct? $sqrt{2 pi n}$ in the numerator?
$endgroup$
– Unknown x
Dec 6 '18 at 17:23
$begingroup$
your alternate approach is very excellent :)
$endgroup$
– Unknown x
Dec 6 '18 at 17:25
|
show 2 more comments
$begingroup$
HINT:
Using Stirling's Formula we have
$$begin{align}
left((n+1)!right)^{1/(n+1)}-left(n!right)^{1/n}&=left(left(sqrt{2pi(n+1)}left(frac{n+1}{e}right)^{n+1}right)left(1+O(1/n)right)right)^{1/(n+1)}\\
&-left(left(sqrt{2pi n}left(frac{n}{e}right)^{n}right)left(1+O(1/n)right)right)^{1/n}
end{align}$$
Can you finish now?
Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
answered Dec 6 '18 at 16:05
Mark ViolaMark Viola
132k1275173
$endgroup$
$begingroup$
This it is not convincing at all. Did you really check that this gets somewhere ?
$endgroup$
– Ewan Delanoy
Dec 6 '18 at 16:13
1
$begingroup$
@EwanDelanoy First of all, it is a HINT only. And yes, I did check this and found that the limit is simply $1/e$.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:15
$begingroup$
@EwanDelanoy Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:22
$begingroup$
Is your sirling formula correct? $sqrt{2 pi n}$ in the numerator?
$endgroup$
– Unknown x
Dec 6 '18 at 17:23
$begingroup$
your alternate approach is very excellent :)
$endgroup$
– Unknown x
Dec 6 '18 at 17:25
|
show 2 more comments
$begingroup$
HINT:
Using Stirling's Formula we have
$$begin{align}
left((n+1)!right)^{1/(n+1)}-left(n!right)^{1/n}&=left(left(sqrt{2pi(n+1)}left(frac{n+1}{e}right)^{n+1}right)left(1+O(1/n)right)right)^{1/(n+1)}\\
&-left(left(sqrt{2pi n}left(frac{n}{e}right)^{n}right)left(1+O(1/n)right)right)^{1/n}
end{align}$$
Can you finish now?
Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
answered Dec 6 '18 at 16:05
Mark ViolaMark Viola
132k1275173
$endgroup$
HINT:
Using Stirling's Formula we have
$$begin{align}
left((n+1)!right)^{1/(n+1)}-left(n!right)^{1/n}&=left(left(sqrt{2pi(n+1)}left(frac{n+1}{e}right)^{n+1}right)left(1+O(1/n)right)right)^{1/(n+1)}\\
&-left(left(sqrt{2pi n}left(frac{n}{e}right)^{n}right)left(1+O(1/n)right)right)^{1/n}
end{align}$$
Can you finish now?
Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
answered Dec 6 '18 at 16:05
Mark ViolaMark Viola
132k1275173
edited Dec 6 '18 at 20:46
answered Dec 6 '18 at 16:05
Mark ViolaMark Viola
132k1275173
answered Dec 6 '18 at 16:05
Mark ViolaMark Viola
132k1275173
answered Dec 6 '18 at 16:05
Mark ViolaMark Viola
132k1275173
132k1275173
$begingroup$
This it is not convincing at all. Did you really check that this gets somewhere ?
$endgroup$
– Ewan Delanoy
Dec 6 '18 at 16:13
1
$begingroup$
@EwanDelanoy First of all, it is a HINT only. And yes, I did check this and found that the limit is simply $1/e$.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:15
$begingroup$
@EwanDelanoy Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:22
$begingroup$
Is your sirling formula correct? $sqrt{2 pi n}$ in the numerator?
$endgroup$
– Unknown x
Dec 6 '18 at 17:23
$begingroup$
your alternate approach is very excellent :)
$endgroup$
– Unknown x
Dec 6 '18 at 17:25
|
show 2 more comments
$begingroup$
This it is not convincing at all. Did you really check that this gets somewhere ?
$endgroup$
– Ewan Delanoy
Dec 6 '18 at 16:13
1
$begingroup$
@EwanDelanoy First of all, it is a HINT only. And yes, I did check this and found that the limit is simply $1/e$.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:15
$begingroup$
@EwanDelanoy Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:22
$begingroup$
Is your sirling formula correct? $sqrt{2 pi n}$ in the numerator?
$endgroup$
– Unknown x
Dec 6 '18 at 17:23
$begingroup$
your alternate approach is very excellent :)
$endgroup$
– Unknown x
Dec 6 '18 at 17:25
$begingroup$
This it is not convincing at all. Did you really check that this gets somewhere ?
$endgroup$
– Ewan Delanoy
Dec 6 '18 at 16:13
$begingroup$
This it is not convincing at all. Did you really check that this gets somewhere ?
$endgroup$
– Ewan Delanoy
Dec 6 '18 at 16:13
1
1
$begingroup$
@EwanDelanoy First of all, it is a HINT only. And yes, I did check this and found that the limit is simply $1/e$.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:15
$begingroup$
@EwanDelanoy First of all, it is a HINT only. And yes, I did check this and found that the limit is simply $1/e$.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:15
$begingroup$
@EwanDelanoy Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:22
$begingroup$
@EwanDelanoy Alternatively, I provided a different, less "brute force" approach in THIS ANSWER.
$endgroup$
– Mark Viola
Dec 6 '18 at 16:22
$begingroup$
Is your sirling formula correct? $sqrt{2 pi n}$ in the numerator?
$endgroup$
– Unknown x
Dec 6 '18 at 17:23
$begingroup$
Is your sirling formula correct? $sqrt{2 pi n}$ in the numerator?
$endgroup$
– Unknown x
Dec 6 '18 at 17:23
$begingroup$
your alternate approach is very excellent :)
$endgroup$
– Unknown x
Dec 6 '18 at 17:25
$begingroup$
your alternate approach is very excellent :)
$endgroup$
– Unknown x
Dec 6 '18 at 17:25
|
show 2 more comments
$begingroup$
Let me check. But the limit is infinity.
$endgroup$
– Unknown x
Dec 6 '18 at 16:01
3
$begingroup$
The limit should be $1/e$.
$endgroup$
– Paramanand Singh
Dec 6 '18 at 16:04
$begingroup$
How do we get?Can you give some hint?
$endgroup$
– Unknown x
Dec 6 '18 at 16:05