Given #a_n=(n!)/n^n# Solve #(a_((n+1)))/(a_n)=?#

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Answer 1 · 2018-06-28T12:52:39

See a solution process below:

Given: #a_n = (n!)/n^n#

Then: #a_(n+1) = ((n + 1)!)/(n + 1)^(n + 1)#

Therefore:

#a_(n+1)/a_n = (((n + 1)!)/(n + 1)^(n + 1))/((n!)/n^n)#

#a_(n+1)/a_n = ((n + 1)!n^n)/(n!(n + 1)^(n + 1))#

#a_(n+1)/a_n = ((n + 1)!)/(n!) xx n^n/((n + 1)^(n + 1))#

#a_(n+1)/a_n = (n!(n + 1))/(n!) xx n^n/((n + 1)^(n + 1))#

#a_(n+1)/a_n = (color(red)(cancel(color(black)(n!)))(n + 1))/color(red)(cancel(color(black)(n!))) xx n^n/((n + 1)^(n + 1))#

#a_(n+1)/a_n = (n + 1) xx n^n/((n + 1)^(n + 1))#

#a_(n+1)/a_n = color(red)(cancel(color(black)((n + 1)))) xx n^n/((n + 1)^(n color(red)(cancel(color(black)(+ 1)))))#

#a_(n+1)/a_n = n^n/((n + 1)^n#

#a_(n+1)/a_n = (n/(n + 1))^n#

Answer 2 · 2018-06-28T12:53:44

The answer is #=(1+1/n)^(-n)#

#a_n=(n!)/(n^n)#

#a_(n+1)=((n+1)!)/((n+1)^(n+1))#

#a_n/a_(n+1)=(n!)/(n^n)*((n+1)^(n+1))/((n+1)!)#

#=1/cancel(n+1)*(cancel(n+1)(n+1)^n)/n^n#

#=(n+1)^n/n^n#

#=((n+1)/n)^n#

#=(1+1/n)^n#.

#:. a_(n+1)/a_n=(1+1/n)^(-n)#.

And

#lim_(n->oo)(1+1/n)^n=e#