sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n)) 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)/(nsqrt(n)) , we may apply Alternating Series Test.
In Alternating Series Test, the series sum (-1)^(n+1) a_n is convergent if:
1) a_ngt=0
2) a_n is monotone and decreasing sequence.
3) lim_(n-gtoo) a_n =0
For the series sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n)) , we have:
a_n = 1/(nsqrt(n))
Apply the radical property: sqrt(x) =x^(1/2) and Law of Exponents: x^n*x^m =x^(n+m).
a_n = 1/(nsqrt(n))
      =1/(n*n^(1/2))
      =1/n^(1+1/2)
      =1/n^(3/2)
The a_n =1/n^(3/2) is a decreasing sequence.
Then, we set-up the limit as :
lim_(n-gtoo)1/n^(3/2) = 1/oo =0
By alternating series test criteria, the series sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n)) converges.
The series sum_(n=1)^oo (-1)^(n+1)/(nsqrt(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=1)^oo |(-1)^(n+1)/(nsqrt(n))| =sum_(n=1)^oo 1/(nsqrt(n))
                         =sum_(n=1)^oo 1/n^(3/2)
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^(3/2) has p=3/2 or 1.5 which satisfies pgt1 . Thus, the sum_(n=1)^oo |(-1)^(n+1)/(nsqrt(n))| is convergent.
Conclusion:
Based on Absolute convergence criteria, the series sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n)) is absolutely convergent since  sum |a_n| as sum_(n=1)^oo |(-1)^(n+1)/(nsqrt(n))| is convergent.