How to prove the divergence of the sequence $(n!)^{1/n}$ [duplicate]
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This question already has an answer here:
Analysis: Prove divergence of sequence $(n!)^{frac2n}$
5 answers
How to prove that
$(n!)^{1/n}$ tends to infinity as $n$ tends to infinity ?
real-analysis sequences-and-series
edited Jan 12 at 12:36
Mohamad Misto
516214
asked Jan 12 at 12:30
Nilesh KhatriNilesh Khatri
195
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marked as duplicate by José Carlos Santos, user416281, Michael Burr, Simply Beautiful Art, Crostul Jan 12 at 13:11
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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$begingroup$
This question already has an answer here:
Analysis: Prove divergence of sequence $(n!)^{frac2n}$
5 answers
How to prove that
$(n!)^{1/n}$ tends to infinity as $n$ tends to infinity ?
real-analysis sequences-and-series
edited Jan 12 at 12:36
Mohamad Misto
516214
asked Jan 12 at 12:30
Nilesh KhatriNilesh Khatri
195
$endgroup$
marked as duplicate by José Carlos Santos, user416281, Michael Burr, Simply Beautiful Art, Crostul Jan 12 at 13:11
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:
Analysis: Prove divergence of sequence $(n!)^{frac2n}$
5 answers
How to prove that
$(n!)^{1/n}$ tends to infinity as $n$ tends to infinity ?
real-analysis sequences-and-series
edited Jan 12 at 12:36
Mohamad Misto
516214
asked Jan 12 at 12:30
Nilesh KhatriNilesh Khatri
195
$endgroup$
This question already has an answer here:
Analysis: Prove divergence of sequence $(n!)^{frac2n}$
5 answers
How to prove that
$(n!)^{1/n}$ tends to infinity as $n$ tends to infinity ?
This question already has an answer here:
Analysis: Prove divergence of sequence $(n!)^{frac2n}$
5 answers
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Jan 12 at 12:36
Mohamad Misto
516214
asked Jan 12 at 12:30
Nilesh KhatriNilesh Khatri
195
edited Jan 12 at 12:36
Mohamad Misto
516214
asked Jan 12 at 12:30
Nilesh KhatriNilesh Khatri
195
edited Jan 12 at 12:36
Mohamad Misto
516214
edited Jan 12 at 12:36
Mohamad Misto
516214
edited Jan 12 at 12:36
Mohamad Misto
516214
516214
asked Jan 12 at 12:30
Nilesh KhatriNilesh Khatri
195
asked Jan 12 at 12:30
Nilesh KhatriNilesh Khatri
195
asked Jan 12 at 12:30
Nilesh KhatriNilesh Khatri
195
195
marked as duplicate by José Carlos Santos, user416281, Michael Burr, Simply Beautiful Art, Crostul Jan 12 at 13:11
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 José Carlos Santos, user416281, Michael Burr, Simply Beautiful Art, Crostul Jan 12 at 13:11
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 |
add a comment |
2 Answers
2
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oldest
votes
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Observe that
$$
n!>left(frac{n}{2}right)^{frac{n}{2}}.
$$
This observation follows from writing out $n!$ as $n(n-1)(n-2)dots2cdot 1$ and taking half of the factors, those of size at least $frac{n}{2}$.
Therefore,
$$
(n!)^{frac{1}{n}}>left(left(frac{n}{2}right)^{frac{n}{2}}right)^{frac{1}{n}}=left(frac{n}{2}right)^{frac{1}{2}},
$$
which diverges.
answered Jan 12 at 12:36
Michael BurrMichael Burr
27k23262
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add a comment |
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Let $k$ be any positive integer. $(n!)^{1/n} geq (k(k+1)cdots n)^{1/n} geq k^{(n-k+1)/n}$ for $n >k$. Can you see from this that $(n!)^{1/n} to infty$?.
answered Jan 12 at 12:35
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Observe that
$$
n!>left(frac{n}{2}right)^{frac{n}{2}}.
$$
This observation follows from writing out $n!$ as $n(n-1)(n-2)dots2cdot 1$ and taking half of the factors, those of size at least $frac{n}{2}$.
Therefore,
$$
(n!)^{frac{1}{n}}>left(left(frac{n}{2}right)^{frac{n}{2}}right)^{frac{1}{n}}=left(frac{n}{2}right)^{frac{1}{2}},
$$
which diverges.
answered Jan 12 at 12:36
Michael BurrMichael Burr
27k23262
$endgroup$
add a comment |
$begingroup$
Observe that
$$
n!>left(frac{n}{2}right)^{frac{n}{2}}.
$$
This observation follows from writing out $n!$ as $n(n-1)(n-2)dots2cdot 1$ and taking half of the factors, those of size at least $frac{n}{2}$.
Therefore,
$$
(n!)^{frac{1}{n}}>left(left(frac{n}{2}right)^{frac{n}{2}}right)^{frac{1}{n}}=left(frac{n}{2}right)^{frac{1}{2}},
$$
which diverges.
answered Jan 12 at 12:36
Michael BurrMichael Burr
27k23262
$endgroup$
add a comment |
$begingroup$
Observe that
$$
n!>left(frac{n}{2}right)^{frac{n}{2}}.
$$
This observation follows from writing out $n!$ as $n(n-1)(n-2)dots2cdot 1$ and taking half of the factors, those of size at least $frac{n}{2}$.
Therefore,
$$
(n!)^{frac{1}{n}}>left(left(frac{n}{2}right)^{frac{n}{2}}right)^{frac{1}{n}}=left(frac{n}{2}right)^{frac{1}{2}},
$$
which diverges.
answered Jan 12 at 12:36
Michael BurrMichael Burr
27k23262
$endgroup$
Observe that
$$
n!>left(frac{n}{2}right)^{frac{n}{2}}.
$$
This observation follows from writing out $n!$ as $n(n-1)(n-2)dots2cdot 1$ and taking half of the factors, those of size at least $frac{n}{2}$.
Therefore,
$$
(n!)^{frac{1}{n}}>left(left(frac{n}{2}right)^{frac{n}{2}}right)^{frac{1}{n}}=left(frac{n}{2}right)^{frac{1}{2}},
$$
which diverges.
answered Jan 12 at 12:36
Michael BurrMichael Burr
27k23262
answered Jan 12 at 12:36
Michael BurrMichael Burr
27k23262
answered Jan 12 at 12:36
Michael BurrMichael Burr
27k23262
answered Jan 12 at 12:36
Michael BurrMichael Burr
27k23262
27k23262
add a comment |
add a comment |
$begingroup$
Let $k$ be any positive integer. $(n!)^{1/n} geq (k(k+1)cdots n)^{1/n} geq k^{(n-k+1)/n}$ for $n >k$. Can you see from this that $(n!)^{1/n} to infty$?.
answered Jan 12 at 12:35
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
$endgroup$
add a comment |
$begingroup$
Let $k$ be any positive integer. $(n!)^{1/n} geq (k(k+1)cdots n)^{1/n} geq k^{(n-k+1)/n}$ for $n >k$. Can you see from this that $(n!)^{1/n} to infty$?.
answered Jan 12 at 12:35
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
$endgroup$
add a comment |
$begingroup$
Let $k$ be any positive integer. $(n!)^{1/n} geq (k(k+1)cdots n)^{1/n} geq k^{(n-k+1)/n}$ for $n >k$. Can you see from this that $(n!)^{1/n} to infty$?.
answered Jan 12 at 12:35
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
$endgroup$
Let $k$ be any positive integer. $(n!)^{1/n} geq (k(k+1)cdots n)^{1/n} geq k^{(n-k+1)/n}$ for $n >k$. Can you see from this that $(n!)^{1/n} to infty$?.
answered Jan 12 at 12:35
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
answered Jan 12 at 12:35
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
answered Jan 12 at 12:35
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
answered Jan 12 at 12:35
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
66.9k53067
add a comment |
add a comment |
