How to prove $lim_{ntoinfty}frac{1}{sqrt[n]{n!}}=0 $ [duplicate]












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  • 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







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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




    $begingroup$
    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
















1












$begingroup$



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







share|cite|improve this question









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marked as duplicate by Martin R, Masacroso, Did, egreg limits
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 Dec 13 '18 at 14:11


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  • 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














1












1








1





$begingroup$



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







share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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:
$$
lim_{ntoinfty} {1over sqrt[n]{n!}} = lim_{ntoinfty} {1over sqrt[n]{2pi n} }left(eover nright) = 1cdot 0 = 0
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, that worked. We didn’t have stirlinga approximation so thanks for the alternative.
    $endgroup$
    – lthz
    Dec 18 '18 at 15:53


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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:
$$
lim_{ntoinfty} {1over sqrt[n]{n!}} = lim_{ntoinfty} {1over sqrt[n]{2pi n} }left(eover nright) = 1cdot 0 = 0
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, that worked. We didn’t have stirlinga approximation so thanks for the alternative.
    $endgroup$
    – lthz
    Dec 18 '18 at 15:53
















2












$begingroup$

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:
$$
lim_{ntoinfty} {1over sqrt[n]{n!}} = lim_{ntoinfty} {1over sqrt[n]{2pi n} }left(eover nright) = 1cdot 0 = 0
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, that worked. We didn’t have stirlinga approximation so thanks for the alternative.
    $endgroup$
    – lthz
    Dec 18 '18 at 15:53














2












2








2





$begingroup$

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:
$$
lim_{ntoinfty} {1over sqrt[n]{n!}} = lim_{ntoinfty} {1over sqrt[n]{2pi n} }left(eover nright) = 1cdot 0 = 0
$$






share|cite|improve this answer









$endgroup$



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:
$$
lim_{ntoinfty} {1over sqrt[n]{n!}} = lim_{ntoinfty} {1over sqrt[n]{2pi n} }left(eover nright) = 1cdot 0 = 0
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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



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