How do you test the alternating series #Sigma (n(-1)^(n+1))/lnn# from n is #[2,oo)# for convergence?

Question

1721 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-09T00:41:27

The series:

#sum_(n=2)^oo (n(-1)^(n+1))/lnn#

is not convergent.

A necessary condition for any series to converge is that:

#lim_(n->oo) a_n = 0#

which also implies:

#lim_(n->oo) abs(a_n) = 0#

In our case:

#lim_(n->oo) abs(a_n) = lim_(n->oo) n/lnn = oo#

so the series is not convergent.

We can also see that taking only the terms of even order:

#lim_(n->oo) a_(2n) = lim_(n->oo) (2n(-1)^(2n+1))/ln(2n) = lim_(n->oo) -(2n)/ln(2n) = -oo#

while for terms of odd order:

#lim_(n->oo) a_(2n+1) = lim_(n->oo) ((2n+1)(-1)^(2n+2))/ln(2n+1) = lim_(n->oo) (2n+1)/ln(2n+1) = +oo#

so the series is irregular.