sum_(n=1)^oo (-1)^n/sqrt(n) Determine the convergence or divergence of the series.

To determine the convergence or divergence of the series sum_(n=1)^oo (-1)^n/sqrt(n) , we may apply the Root Test.
In Root test, we determine the limit as:
lim_(n-gtoo) root(n)(|a_n|)= L
or
lim_(n-gtoo) |a_n|^(1/n)= L
 Then ,we follow the conditions:
a) L lt1 then the series converges absolutely.
b) Lgt1 then the series diverges.
c) L=1 or does not exist  then the test is inconclusive.The series may be divergent, conditionally convergent, or absolutely convergent.
For the given series sum_(n=1)^oo (-1)^n/sqrt(n) , we have a_n =(-1)^n/sqrt(n) .
Applying the Root test, we set-up the limit as: 
lim_(n-gtoo) |(-1)^n/sqrt(n)|^(1/n) =lim_(n-gtoo) (1/sqrt(n))^(1/n)
Note: |(-1)^n| = 1
Apply radical property: root(n)(x) =x^(1/n) and Law of exponent: (x/y)^n = x^n/y^n.
lim_(n-gtoo) (1/sqrt(n))^(1/n) =lim_(n-gtoo) (1/n^(1/2))^(1/n)
                          =lim_(n-gtoo) 1^(1/n) /n^(1/2*1/n)
                          =lim_(n-gtoo) 1^(1/n) /n^(1/(2n))
                          =lim_(n-gtoo) 1 /n^(1/(2n))
Apply the limit property: lim_(x-gta)[(f(x))/(g(x))] =(lim_(x-gta) f(x))/(lim_(x-gta) g(x)).
lim_(n-gtoo) 1 /n^(1/(2n)) =(lim_(n-gtoo) 1 )/(lim_(n-gtoo)n^(1/(2n)))
                      = 1/1
                      =1
The limit value L = 1 implies that the series may be divergent, conditionally convergent, or absolutely convergent.
To verify, we use alternating series test on sum (-1)^n a_n .
a_n = 1/sqrt(n) is positive and decreasing from N=1 .
lim_(n-gtoo)1/sqrt(n) = 1/oo = 0
Based on alternating series test condition,  the series sum_(n=1)^oo (-1)^n/sqrt(n) converges.
Apply p-series test on sum |a_n| .
sum_(n=1)^oo |(-1)^n/sqrt(n)|=sum_(n=1)^oo 1/sqrt(n)
                      =sum_(n=1)^oo 1/n^(1/2)
Based on p-series test condition,  we have p=1/2 that satisfies 0ltplt=1 .
Thus, the series  sum_(n=1)^oo |(-1)^n/sqrt(n)| diverges.      
Notes:
In p-series test, we follow:
-if p is within the interval of 0ltplt=1 then the series diverges.
-if p is within the interval of pgt1 then the series converges.
Absolute Convergence:  sum a_n  is absolutely convergent if sum|a_n|   is convergent.             
Conditional Convergence:  sum a_n is conditionally convergent if sum|a_n|  is divergent and sum a_n is convergent.       
              
Conclusion:
The series sum_(n=1)^oo (-1)^n/sqrt(n) is conditionally convergent since  sum_(n=1)^oo (-1)^n/sqrt(n) is convergent  and sum_(n=1)^oo |(-1)^n/sqrt(n)| is divergent.