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

To determine the convergence or divergence of the series sum_(n=1)^oo (-1)^(n+1)/n^2 , we may apply Alternating Series Test.
In Alternating Series Test, the series sum (-1)^(n+1)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=1)^oo (-1)^(n+1)/n^2 , we have:
a_n = 1/(n^2) which is a decreasing sequence.
As "n " increases, the 1/n^2 decreases.
Then, we set-up the limit as :
lim_(n-gtoo)1/n^2 = 1/oo =0
By alternating series test criteria, the seriessum_(n=1)^oo (-1)^(n+1)/n^2 converges.
The series sum_(n=1)^oo (-1)^(n+1)/n^2 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=1)^oo |(-1)^(n+1)/n^2| =sum_(n=1)^oo 1/n^2
Apply the p-series test sum_(n=1)^oo 1/n^p is convergent if pgt1 and divergent if 0ltplt=1 .
The series sum_(n=1)^oo 1/n^2 has p=2 which satisfies pgt1 . Thus, the series sum_(n=1)^oo |(-1)^(n+1)/n^2| is convergent.
 
Conclusion:
The series sum_(n=1)^oo (-1)^(n+1)/n^2 is absolutely convergent since sum |a_n| as sum_(n=1)^oo |(-1)^(n+1)/n^2| is convergent.