#lim_(n->oo)(1/n((n+1)(n+2)(n+3)cdots(2n))^(1/n))?#

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Answer 1 · 2016-12-13T17:14:17

#4/e#

we have

#((n+1)(n+2)(n+3)cdots(2n)) =((2n)!)/(n!)#

#log_e(n!) approx nlog_en-n# in

#y = 1/n(((2n)!)/(n!))^(1/n)#

we have

#y = 1/n(((2n)^(2n)e^(-2n))/(n^n e^(-n)))^(1/n)=4/e#

So the answer is

#lim_(n->oo)(1/n((n+1)(n+2)(n+3)cdots(2n))^(1/n))=4/e#