How to prove $lim_{ntoinfty}frac{1}{sqrt[n]{n!}}=0 $ [duplicate]
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This question already has an answer here:
Prove $limlimits_{ntoinfty}frac{1}{sqrt[n]{n!}}=0$
7 answers
I need to prove the limit of:
$$lim_{ntoinfty}frac{1}{sqrt[n]{n!}}=0 $$
I have tried bounding it by a pairwise bigger sequence but I cannot find one that is suitable. I also cannot figure out an $epsilon$-proof, since I don't know how to adequately handle the factorial to get a substitution for n such that the convergence criterion is fulfilled.
limits analysis
asked Dec 13 '18 at 13:32
lthzlthz
1246
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marked as duplicate by Martin R, Masacroso, Did, egreg limits
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Dec 13 '18 at 14:11
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:
Prove $limlimits_{ntoinfty}frac{1}{sqrt[n]{n!}}=0$
7 answers
I need to prove the limit of:
$$lim_{ntoinfty}frac{1}{sqrt[n]{n!}}=0 $$
I have tried bounding it by a pairwise bigger sequence but I cannot find one that is suitable. I also cannot figure out an $epsilon$-proof, since I don't know how to adequately handle the factorial to get a substitution for n such that the convergence criterion is fulfilled.
limits analysis
asked Dec 13 '18 at 13:32
lthzlthz
1246
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marked as duplicate by Martin R, Masacroso, Did, egreg limits
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Dec 13 '18 at 14:11
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.
4
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Using stirling approximation $n!=bigg(frac{n}{e}bigg)^n sqrt{2pi n}.$
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– DXT
Dec 13 '18 at 13:39
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Arithmetic-Geometric Mean inequality makes short work of this.
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– Theo Bendit
Dec 13 '18 at 13:41
2
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Also: math.stackexchange.com/q/490163, math.stackexchange.com/a/706474, math.stackexchange.com/q/136626,
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– Martin R
Dec 13 '18 at 13:53
add a comment |
$begingroup$
This question already has an answer here:
Prove $limlimits_{ntoinfty}frac{1}{sqrt[n]{n!}}=0$
7 answers
I need to prove the limit of:
$$lim_{ntoinfty}frac{1}{sqrt[n]{n!}}=0 $$
I have tried bounding it by a pairwise bigger sequence but I cannot find one that is suitable. I also cannot figure out an $epsilon$-proof, since I don't know how to adequately handle the factorial to get a substitution for n such that the convergence criterion is fulfilled.
limits analysis
asked Dec 13 '18 at 13:32
lthzlthz
1246
$endgroup$
This question already has an answer here:
Prove $limlimits_{ntoinfty}frac{1}{sqrt[n]{n!}}=0$
7 answers
I need to prove the limit of:
$$lim_{ntoinfty}frac{1}{sqrt[n]{n!}}=0 $$
I have tried bounding it by a pairwise bigger sequence but I cannot find one that is suitable. I also cannot figure out an $epsilon$-proof, since I don't know how to adequately handle the factorial to get a substitution for n such that the convergence criterion is fulfilled.
This question already has an answer here:
Prove $limlimits_{ntoinfty}frac{1}{sqrt[n]{n!}}=0$
7 answers
limits analysis
limits analysis
asked Dec 13 '18 at 13:32
lthzlthz
1246
asked Dec 13 '18 at 13:32
lthzlthz
1246
asked Dec 13 '18 at 13:32
lthzlthz
1246
asked Dec 13 '18 at 13:32
lthzlthz
1246
asked Dec 13 '18 at 13:32
lthzlthz
1246
1246
marked as duplicate by Martin R, Masacroso, Did, egreg limits
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Dec 13 '18 at 14:11
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 Martin R, Masacroso, Did, egreg limits
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Dec 13 '18 at 14:11
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.
4
$begingroup$
Using stirling approximation $n!=bigg(frac{n}{e}bigg)^n sqrt{2pi n}.$
$endgroup$
– DXT
Dec 13 '18 at 13:39
$begingroup$
Arithmetic-Geometric Mean inequality makes short work of this.
$endgroup$
– Theo Bendit
Dec 13 '18 at 13:41
2
$begingroup$
Also: math.stackexchange.com/q/490163, math.stackexchange.com/a/706474, math.stackexchange.com/q/136626,
$endgroup$
– Martin R
Dec 13 '18 at 13:53
add a comment |
4
$begingroup$
Using stirling approximation $n!=bigg(frac{n}{e}bigg)^n sqrt{2pi n}.$
$endgroup$
– DXT
Dec 13 '18 at 13:39
$begingroup$
Arithmetic-Geometric Mean inequality makes short work of this.
$endgroup$
– Theo Bendit
Dec 13 '18 at 13:41
2
$begingroup$
Also: math.stackexchange.com/q/490163, math.stackexchange.com/a/706474, math.stackexchange.com/q/136626,
$endgroup$
– Martin R
Dec 13 '18 at 13:53
4
4
$begingroup$
Using stirling approximation $n!=bigg(frac{n}{e}bigg)^n sqrt{2pi n}.$
$endgroup$
– DXT
Dec 13 '18 at 13:39
$begingroup$
Using stirling approximation $n!=bigg(frac{n}{e}bigg)^n sqrt{2pi n}.$
$endgroup$
– DXT
Dec 13 '18 at 13:39
$begingroup$
Arithmetic-Geometric Mean inequality makes short work of this.
$endgroup$
– Theo Bendit
Dec 13 '18 at 13:41
$begingroup$
Arithmetic-Geometric Mean inequality makes short work of this.
$endgroup$
– Theo Bendit
Dec 13 '18 at 13:41
2
2
$begingroup$
Also: math.stackexchange.com/q/490163, math.stackexchange.com/a/706474, math.stackexchange.com/q/136626,
$endgroup$
– Martin R
Dec 13 '18 at 13:53
$begingroup$
Also: math.stackexchange.com/q/490163, math.stackexchange.com/a/706474, math.stackexchange.com/q/136626,
$endgroup$
– Martin R
Dec 13 '18 at 13:53
add a comment |
1 Answer
1
active
oldest
votes
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Consider the following:
$$
n! = n(n-1)(n-2)cdots 3 cdot 2 cdot 1 geq underbrace{{nover 2}cdot {nover 2}cdots {n over 2}}_{{n over 2}text{times}} = left({nover 2}right)^{nover 2} iff \
iff sqrt[n]{n!} ge left({left({nover 2}right)^{nover 2}}right)^{1over n} = sqrt{nover 2} iff {1over sqrt[n]{n!}} le sqrt{2over n}
$$
Also $x_n ge 0$, so by squeeze theorem:
$$
0 le lim_{ntoinfty} {1over sqrt[n]{n!}} le lim_{nto infty} sqrt{2over n} = 0
$$
Another way would be Stirling's approximation. Recall:
$$
n! sim sqrt{2pi n} left(nover eright)^{n}
$$
So:
$$
lim_{ntoinfty} {1over sqrt[n]{n!}} = lim_{ntoinfty} {1over sqrt[n]{2pi n} }left(eover nright) = 1cdot 0 = 0
$$
answered Dec 13 '18 at 14:06
romanroman
2,18521224
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Thank you, that worked. We didn’t have stirlinga approximation so thanks for the alternative.
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– lthz
Dec 18 '18 at 15:53
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the following:
$$
n! = n(n-1)(n-2)cdots 3 cdot 2 cdot 1 geq underbrace{{nover 2}cdot {nover 2}cdots {n over 2}}_{{n over 2}text{times}} = left({nover 2}right)^{nover 2} iff \
iff sqrt[n]{n!} ge left({left({nover 2}right)^{nover 2}}right)^{1over n} = sqrt{nover 2} iff {1over sqrt[n]{n!}} le sqrt{2over n}
$$
Also $x_n ge 0$, so by squeeze theorem:
$$
0 le lim_{ntoinfty} {1over sqrt[n]{n!}} le lim_{nto infty} sqrt{2over n} = 0
$$
Another way would be Stirling's approximation. Recall:
$$
n! sim sqrt{2pi n} left(nover eright)^{n}
$$
So:
$$
lim_{ntoinfty} {1over sqrt[n]{n!}} = lim_{ntoinfty} {1over sqrt[n]{2pi n} }left(eover nright) = 1cdot 0 = 0
$$
answered Dec 13 '18 at 14:06
romanroman
2,18521224
$endgroup$
$begingroup$
Thank you, that worked. We didn’t have stirlinga approximation so thanks for the alternative.
$endgroup$
– lthz
Dec 18 '18 at 15:53
add a comment |
$begingroup$
Consider the following:
$$
n! = n(n-1)(n-2)cdots 3 cdot 2 cdot 1 geq underbrace{{nover 2}cdot {nover 2}cdots {n over 2}}_{{n over 2}text{times}} = left({nover 2}right)^{nover 2} iff \
iff sqrt[n]{n!} ge left({left({nover 2}right)^{nover 2}}right)^{1over n} = sqrt{nover 2} iff {1over sqrt[n]{n!}} le sqrt{2over n}
$$
Also $x_n ge 0$, so by squeeze theorem:
$$
0 le lim_{ntoinfty} {1over sqrt[n]{n!}} le lim_{nto infty} sqrt{2over n} = 0
$$
Another way would be Stirling's approximation. Recall:
$$
n! sim sqrt{2pi n} left(nover eright)^{n}
$$
So:
$$
lim_{ntoinfty} {1over sqrt[n]{n!}} = lim_{ntoinfty} {1over sqrt[n]{2pi n} }left(eover nright) = 1cdot 0 = 0
$$
answered Dec 13 '18 at 14:06
romanroman
2,18521224
$endgroup$
$begingroup$
Thank you, that worked. We didn’t have stirlinga approximation so thanks for the alternative.
$endgroup$
– lthz
Dec 18 '18 at 15:53
add a comment |
$begingroup$
Consider the following:
$$
n! = n(n-1)(n-2)cdots 3 cdot 2 cdot 1 geq underbrace{{nover 2}cdot {nover 2}cdots {n over 2}}_{{n over 2}text{times}} = left({nover 2}right)^{nover 2} iff \
iff sqrt[n]{n!} ge left({left({nover 2}right)^{nover 2}}right)^{1over n} = sqrt{nover 2} iff {1over sqrt[n]{n!}} le sqrt{2over n}
$$
Also $x_n ge 0$, so by squeeze theorem:
$$
0 le lim_{ntoinfty} {1over sqrt[n]{n!}} le lim_{nto infty} sqrt{2over n} = 0
$$
Another way would be Stirling's approximation. Recall:
$$
n! sim sqrt{2pi n} left(nover eright)^{n}
$$
So:
$$
lim_{ntoinfty} {1over sqrt[n]{n!}} = lim_{ntoinfty} {1over sqrt[n]{2pi n} }left(eover nright) = 1cdot 0 = 0
$$
answered Dec 13 '18 at 14:06
romanroman
2,18521224
$endgroup$
Consider the following:
$$
n! = n(n-1)(n-2)cdots 3 cdot 2 cdot 1 geq underbrace{{nover 2}cdot {nover 2}cdots {n over 2}}_{{n over 2}text{times}} = left({nover 2}right)^{nover 2} iff \
iff sqrt[n]{n!} ge left({left({nover 2}right)^{nover 2}}right)^{1over n} = sqrt{nover 2} iff {1over sqrt[n]{n!}} le sqrt{2over n}
$$
Also $x_n ge 0$, so by squeeze theorem:
$$
0 le lim_{ntoinfty} {1over sqrt[n]{n!}} le lim_{nto infty} sqrt{2over n} = 0
$$
Another way would be Stirling's approximation. Recall:
$$
n! sim sqrt{2pi n} left(nover eright)^{n}
$$
So:
$$
lim_{ntoinfty} {1over sqrt[n]{n!}} = lim_{ntoinfty} {1over sqrt[n]{2pi n} }left(eover nright) = 1cdot 0 = 0
$$
answered Dec 13 '18 at 14:06
romanroman
2,18521224
answered Dec 13 '18 at 14:06
romanroman
2,18521224
answered Dec 13 '18 at 14:06
romanroman
2,18521224
answered Dec 13 '18 at 14:06
romanroman
2,18521224
2,18521224
$begingroup$
Thank you, that worked. We didn’t have stirlinga approximation so thanks for the alternative.
$endgroup$
– lthz
Dec 18 '18 at 15:53
add a comment |
$begingroup$
Thank you, that worked. We didn’t have stirlinga approximation so thanks for the alternative.
$endgroup$
– lthz
Dec 18 '18 at 15:53
$begingroup$
Thank you, that worked. We didn’t have stirlinga approximation so thanks for the alternative.
$endgroup$
– lthz
Dec 18 '18 at 15:53
$begingroup$
Thank you, that worked. We didn’t have stirlinga approximation so thanks for the alternative.
$endgroup$
– lthz
Dec 18 '18 at 15:53
add a comment |
4
$begingroup$
Using stirling approximation $n!=bigg(frac{n}{e}bigg)^n sqrt{2pi n}.$
$endgroup$
– DXT
Dec 13 '18 at 13:39
$begingroup$
Arithmetic-Geometric Mean inequality makes short work of this.
$endgroup$
– Theo Bendit
Dec 13 '18 at 13:41
2
$begingroup$
Also: math.stackexchange.com/q/490163, math.stackexchange.com/a/706474, math.stackexchange.com/q/136626,
$endgroup$
– Martin R
Dec 13 '18 at 13:53