Given the series:
sum_(n=1)^oo (-1)^n x^n/n
we can use the ratio test by evaluating:
abs(a_(n+1)/a_n) = abs ( ( (-1)^(n+1) x^(n+1) /(n+1) ) / ( (-1^n) x^n/n)) = n/(n+1) absx
So that the ratio limit is:
lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) n/(n+1) absx = abs x
meaning that for abs(x) < 1 the series is absolutely convergent and for abs(x) > 1 it is not convergent.
For abs x = 1 we have two cases:
(i) x = 1
sum_(n=1)^oo (-1)^n x^n/n = sum_(n=1)^oo (-1)^n/n which is the alternate harmonic series and is convergent.
(ii) x = -1
sum_(n=1)^oo (-1)^n x^n/n = sum_(n=1)^oo (-1)^n(-1)^n/n = sum_(n=1)^oo 1/n which is the harmonic series and is divergent.
In conclusion the series is convergent for x in (-1,1]