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)