How do you apply the ratio test to determine if #Sigma n^n/((2n)!)# from #n=[1,oo)# is convergent to divergent?

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Answer 1 · 2017-03-14T13:32:15

The series:

#sum_(n=1)^oo n^n/((2n)!)#

is convergent.

Evaluate the ratio:

#abs (a_(n+1)/a_n) = ((n+1)^(n+1)/((2(n+1)!)) )/ (n^n/((2n)!)) #

#abs (a_(n+1)/a_n) = (n+1)^(n+1)/n^n ((2n)!)/((2(n+1)!)) #

#abs (a_(n+1)/a_n) = (n+1)((n+1)/n)^n ((2n)!)/((2n+2)!) #

#abs (a_(n+1)/a_n) = (n+1)(1+1/n)^n 1/((2n+2)(2n+1)) #

#abs (a_(n+1)/a_n) =(1+1/n)^n cancel(n+1) /(2cancel((n+1))(2n+1)) #

#abs (a_(n+1)/a_n) =(1+1/n)^n 1 /(2(2n+1)) #

Now we have:

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

#lim_(n->oo) 1/(2n+1) = 0#

so:

#lim_(n->oo) abs (a_(n+1)/a_n) = 0#

which proves the series to be convergent based on the ratio test.