sum_(n=1)^oo (-1)^nx^n/n Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.)

To determine the interval of convergence for the given series: sum_(n=1)^oo(-1)^nx^n/n , we may apply 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
The series is  absolutely convergent if it satisfies the Root test condition: L lt1.
For the  given series: sum_(n=1)^oo(-1)^nx^n/n , we have:
a_n= (-1)^nx^n/n
 Then, set-up the limit as:
lim_(n-gtoo) |(-1)^nx^n/n|^(1/n) =lim_(n-gtoo) |x^n/n|^(1/n)
Note: |(-1)^n|=1 and 1*|x^n| =|x^n| .
Apply Law of exponents: (x/y)^n =x^n/y^n and (x^n)^m= x^(n*m).
lim_(n-gtoo) |x^n/n|^(1/n)=lim_(n-gtoo) |(x^n)^(1/n)/n^(1/n)|
                        =lim_(n-gtoo) |x^(n*1/n)/n^(1/n)|
                        =lim_(n-gtoo) |x^(n/n)/n^(1/n)|
                         =lim_(n-gtoo) |x^1/n^(1/n)|
                         =lim_(n-gtoo) |x/n^(1/n)|
Evaluate the limit.
lim_(n-gtoo) |x/n^(1/n)| =( lim_(n-gtoo) |x|)/( lim_(n-gtoo) |n^(1/n)|)
                     = |x|/|1|
                     = |x|
Thus, we have the limit value: L = |x| .
 To determine the interval of convergence , we plug-in L=|x| on the condition for convergent series: Llt1 .
|x| lt 1
-1ltxlt1
 The series may converge at L=1 . To verify this, we check for the possible convergence at the endpoints of the interval of x.
Using x=-1 on the series sum_(n=1)^oo(-1)^nx^n/n , we get: 
sum_(n=1)^oo(-1)^n(-1)^n/n=sum_(n=1)^oo(-1*-1)^n/n
                              =sum_(n=1)^oo 1^n/n
                              =sum_(n=1)^oo 1/n
The sum_(n=1)^oo 1/n is in a form of a p-series where p =1  satisfies 0ltplt=1 . According to p-series test of convergence, the series sum_(n=1)^oo 1/n^p is convergent if pgt1 and divergent if  0ltplt=1 .
Thus, the series sum_(n=1)^oo 1/n  is divergent at x=-1 .
Using x=1 on the series sum_(n=1)^oo(-1)^nx^n/n , we get: 
sum_(n=1)^oo(-1)^n1^n/n=sum_(n=1)^oo(-1*1)^n/n
                           =sum_(n=1)^oo (-1)^n/n
Applying alternating series test on the series sum (-1)^na_n where a_ngt=0 for all n , the series is convergent if we have:
1. lim_(n-gtoo) a_n = 0
2. a_n is a decreasing sequence
In the series sum_(n=1)^oo (-1)^n/n or sum_(n=1)^oo (-1)^n 1/n , the a_n= 1/n is decreasing sequence and 1/ngt=0 for all n .
Evaluating the limit:
lim_(n-gtoo) 1/n = 1/oo = 0
Thus,  sum_(n=1)^oo (-1)^n/n is convergent  at x=1 .
The series sum_(n=1)^oo (-1)^n/n has a positive and negative elements. We check for absolute or conditional convergence.
The sum a_n as sum_(n=1)^oo (-1)^n/n  is convergent based from alternating series criteria.
 The sum |a_n| as sum_(n=1)^oo |(-1)^n/n|  or  sum_(n=1)^oo 1/n is divergent based form p-series criteria.
 Thus, the series sum_(n=1)^oo (-1)^n/n is conditionally convergent at x=1 .
Therefore, the power series sum_(n=1)^oo(-1)^nx^n/n  has an interval of convergence: -1ltxlt=1 .