Find $lim_n to inftyleft(frac(3n)!(2n)!n^nright)^1/n$ [closed]

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Find $$lim_n to inftyleft(frac(3n)!(2n)!n^nright)^1/n$$




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closed as off-topic by user223391, Batominovski, Siong Thye Goh, José Carlos Santos, heropup Jul 30 at 7:54


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    Please see math.meta.stackexchange.com/questions/5020
    – Lord Shark the Unknown
    Jul 30 at 6:44














up vote
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Find $$lim_n to inftyleft(frac(3n)!(2n)!n^nright)^1/n$$




sequences-and-series




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closed as off-topic by user223391, Batominovski, Siong Thye Goh, José Carlos Santos, heropup Jul 30 at 7:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Batominovski, Siong Thye Goh, José Carlos Santos, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    Please see math.meta.stackexchange.com/questions/5020
    – Lord Shark the Unknown
    Jul 30 at 6:44












up vote
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up vote
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down vote

favorite











Find $$lim_n to inftyleft(frac(3n)!(2n)!n^nright)^1/n$$




sequences-and-series




share|cite|improve this question













Find $$lim_n to inftyleft(frac(3n)!(2n)!n^nright)^1/n$$





sequences-and-series





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 30 at 6:54









Bill O'Haran

2,4411418




2,4411418









asked Jul 30 at 6:38









Afif bakr

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11




closed as off-topic by user223391, Batominovski, Siong Thye Goh, José Carlos Santos, heropup Jul 30 at 7:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Batominovski, Siong Thye Goh, José Carlos Santos, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by user223391, Batominovski, Siong Thye Goh, José Carlos Santos, heropup Jul 30 at 7:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Batominovski, Siong Thye Goh, José Carlos Santos, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    Please see math.meta.stackexchange.com/questions/5020
    – Lord Shark the Unknown
    Jul 30 at 6:44












  • 1




    Please see math.meta.stackexchange.com/questions/5020
    – Lord Shark the Unknown
    Jul 30 at 6:44







1




1




Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 30 at 6:44




Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 30 at 6:44










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Given $$lim_nrightarrow inftybigg[frac(3n)!(2n)!cdot n^nbigg]^frac1n$$



Using stirling approximation $$n!=bigg(fracnebigg)^frac1nsqrt2pi n$$



So $$lim_nrightarrow inftyBigg[fracbigg(frac3nebigg)^3nsqrt6pi nbigg(frac2nebigg)^2nsqrt4pi ncdot n^nBigg]^frac1n.$$



So $$lim_nrightarrow inftyfrac274ebigg(fracsqrt6pi nsqrt2pi nbigg)^frac1n=frac274e.$$






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    You mean: $enspacedisplaystyle n!approxleft(fracneright)^nsqrt2pi nenspace$
    – user90369
    Jul 30 at 7:29











  • Some $e$ got lost in the last passage.
    – Saucy O'Path
    Jul 30 at 7:53


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













Given $$lim_nrightarrow inftybigg[frac(3n)!(2n)!cdot n^nbigg]^frac1n$$



Using stirling approximation $$n!=bigg(fracnebigg)^frac1nsqrt2pi n$$



So $$lim_nrightarrow inftyBigg[fracbigg(frac3nebigg)^3nsqrt6pi nbigg(frac2nebigg)^2nsqrt4pi ncdot n^nBigg]^frac1n.$$



So $$lim_nrightarrow inftyfrac274ebigg(fracsqrt6pi nsqrt2pi nbigg)^frac1n=frac274e.$$






share|cite|improve this answer



















  • 1




    You mean: $enspacedisplaystyle n!approxleft(fracneright)^nsqrt2pi nenspace$
    – user90369
    Jul 30 at 7:29











  • Some $e$ got lost in the last passage.
    – Saucy O'Path
    Jul 30 at 7:53















up vote
2
down vote













Given $$lim_nrightarrow inftybigg[frac(3n)!(2n)!cdot n^nbigg]^frac1n$$



Using stirling approximation $$n!=bigg(fracnebigg)^frac1nsqrt2pi n$$



So $$lim_nrightarrow inftyBigg[fracbigg(frac3nebigg)^3nsqrt6pi nbigg(frac2nebigg)^2nsqrt4pi ncdot n^nBigg]^frac1n.$$



So $$lim_nrightarrow inftyfrac274ebigg(fracsqrt6pi nsqrt2pi nbigg)^frac1n=frac274e.$$






share|cite|improve this answer



















  • 1




    You mean: $enspacedisplaystyle n!approxleft(fracneright)^nsqrt2pi nenspace$
    – user90369
    Jul 30 at 7:29











  • Some $e$ got lost in the last passage.
    – Saucy O'Path
    Jul 30 at 7:53













up vote
2
down vote










up vote
2
down vote









Given $$lim_nrightarrow inftybigg[frac(3n)!(2n)!cdot n^nbigg]^frac1n$$



Using stirling approximation $$n!=bigg(fracnebigg)^frac1nsqrt2pi n$$



So $$lim_nrightarrow inftyBigg[fracbigg(frac3nebigg)^3nsqrt6pi nbigg(frac2nebigg)^2nsqrt4pi ncdot n^nBigg]^frac1n.$$



So $$lim_nrightarrow inftyfrac274ebigg(fracsqrt6pi nsqrt2pi nbigg)^frac1n=frac274e.$$






share|cite|improve this answer















Given $$lim_nrightarrow inftybigg[frac(3n)!(2n)!cdot n^nbigg]^frac1n$$



Using stirling approximation $$n!=bigg(fracnebigg)^frac1nsqrt2pi n$$



So $$lim_nrightarrow inftyBigg[fracbigg(frac3nebigg)^3nsqrt6pi nbigg(frac2nebigg)^2nsqrt4pi ncdot n^nBigg]^frac1n.$$



So $$lim_nrightarrow inftyfrac274ebigg(fracsqrt6pi nsqrt2pi nbigg)^frac1n=frac274e.$$







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 30 at 9:29


























answered Jul 30 at 6:52









Durgesh Tiwari

4,7782426




4,7782426







  • 1




    You mean: $enspacedisplaystyle n!approxleft(fracneright)^nsqrt2pi nenspace$
    – user90369
    Jul 30 at 7:29











  • Some $e$ got lost in the last passage.
    – Saucy O'Path
    Jul 30 at 7:53













  • 1




    You mean: $enspacedisplaystyle n!approxleft(fracneright)^nsqrt2pi nenspace$
    – user90369
    Jul 30 at 7:29











  • Some $e$ got lost in the last passage.
    – Saucy O'Path
    Jul 30 at 7:53








1




1




You mean: $enspacedisplaystyle n!approxleft(fracneright)^nsqrt2pi nenspace$
– user90369
Jul 30 at 7:29





You mean: $enspacedisplaystyle n!approxleft(fracneright)^nsqrt2pi nenspace$
– user90369
Jul 30 at 7:29













Some $e$ got lost in the last passage.
– Saucy O'Path
Jul 30 at 7:53





Some $e$ got lost in the last passage.
– Saucy O'Path
Jul 30 at 7:53



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