How do you test the series #Sigma n^-n# from n is #[1,oo)# for convergence?

1 Answer
Jun 4, 2017

Using the ratio test

Explanation:

The ratio test finds the ratio between terms #u_k and u_(k+1)# as #k# tends to infinity, and if #abs(u_(k+1)/u_k)<1 (k->oo)#, then the series is found to be convergent, as it means that terms are getting progressively smaller and tending towards 0.

Hence, in order to test #sum n^-n, n in [1,oo)# for convergence, we find the ratio #abs(u_(k+1)/u_k)(k->oo)#:

#k->oo abs(u_(k+1)/u_k)#
#= k->ooabs((k+1)^-(k+1)/k^-k)#
#=k->oo abs(k^k/(k+1)^(k+1))#
#=k->oo abs(k^k/(k+1)^k*1/(k+1))#
#=k->oo abs((k/(k+1))^k*1/(k+1))#
#=k->oo abs((1/(1+1/k))^k*1/(k+1))#
#because k->oo 1/k=0#
#therefore k->ooabs((1/(1+1/k))^k)=k->ooabs((1/(1+0))^k)=abs(1)=1#
#therefore k->ooabs((1/(1+1/k))^k*1/(k+1))=k->ooabs(1*1/(k+1))#

#=k->ooabs(1/(k+1))=0#

#because k->oo abs(u_(k+1)/u_k)=0#
i.e. #k->oo abs(u_(k+1)/u_k)<1#
#therefore# By the ratio test, the series #sum n^-n# converges from #n in [1,oo)#