How you solve this ?#lim_(n->oo)(5^(n)n!)/(2^(n)n^n)#

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Answer 1 · 2017-03-20T23:23:03

#lim_(n->oo)(5^(n)n!)/(2^(n)n^n)=0#

Using Stirling assymptotic approximation

#n! approx sqrt(2pi n)(n/e)^n# we have

#(n!)/n^n approx sqrt(2pi n)e^(-n)# so

#lim_(n->oo)(5^(n)n!)/(2^(n)n^n)=(5/2)^n sqrt(2pi n)e^(-n) = (5/(2e))^n sqrt(2pi n) = a^n sqrt(2pi n)# with #a < 1#

then

#lim_(n->oo)(5^(n)n!)/(2^(n)n^n)=0#