Question

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

Answer 1

The series:

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

is not convergent.

Explanation:

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.