sum_(n=2)^oo (-1)^n/(nlnn) Determine whether the series converges absolutely or conditionally, or diverges.

To determine the convergence or divergence of the series sum_(n=2)^oo (-1)^n/(nln(n)) , we may apply Alternating Series Test.
In Alternating Series Test, the series sum (-1)^n a_n is convergent if:
1) a_n is monotone and decreasing sequence.
2) lim_(n-gtoo) a_n =0
3) a_ngt=0
For the series sum_(n=2)^oo (-1)^n/(nln(n)) , we have:
a_n = 1/(nln(n)) which is a positive, continuous, and decreasing sequence from N=2.
Note: As "n " increases, the nln(n) increases then 1/(nln(n)) decreases.
Then, we set-up the limit as :
lim_(n-gtoo)1/(nln(n))= 1/oo =0
By alternating series test criteria, the series sum_(n=2)^oo (-1)^n/(nln(n))  converges.
The series sum_(n=2)^oo (-1)^n/(nln(n))  has positive and negative elements. Thus, we must verify if the series converges absolutely or conditionally. Recall:
a) Absolute Convergence:  sum a_n  is absolutely convergent if sum|a_n|   is convergent.  
b) Conditional Convergence:  sum a_n is conditionally convergent if sum|a_n|  is divergent and sum a_n  is convergent.  
We evaluate the sum |a_n| as :
sum_(n=2)^oo |(-1)^n/(nln(n))|=sum_(n=2)^oo 1/(nln(n))
Applying integral test for convergence, we evaluate the series as:
int_2^oo1/(nln(n))dn=lim_(n-gtoo) int_2^t 1/(nln(n))dn
Apply u-substitution: u =ln(n) then du =1/ndn .
int 1/(nln(n))dn =int 1/(ln(n))*1/ndn
                       =int 1/u du
                        =ln|u|
Plug-in u=ln(n) on the indefinite integral ln|u| , we get:
int_2^t 1/(nln(n))dn =ln|ln(n)||_2^t
Applying definite integral formula: F(x)|_a^b = F(b)-F(a) .
ln|ln(n)||_2^t =ln|ln(t)|-ln|ln(2)|
Then, the limit becomes:
lim_(n-gtoo) int_2^t1/(nln(n))dn =lim_(n-gtoo) [ln|ln(t)|-ln|ln(2)|]
                                   =lim_(n-gtoo)ln|ln(t)|-lim_(n-gtoo)ln|ln(2)|
                                   = oo - ln|ln(2 )|
                                   =oo
int_2^oo1/(nln(n))dn=oo implies the series  sum_(n=2)^oo |(-1)^n/(nln(n))| diverges.
 
Conclusion:  
The series sum_(n=2)^oo (-1)^n/(nln(n)) is conditionally convergent sincesum |a_n| as   sum_(n=2)^oo |(-1)^n/(nln(n))| is divergent and sum a_n as sum_(n=2)^oo (-1)^n/(nln(n)) is convergent.