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

Given a_n=(n!)/n^n

Solve (a_((n+1)))/(a_n)=?

2 Answers
Jun 28, 2018

See a solution process below:

Explanation:

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

Jun 28, 2018

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

Explanation:

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