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]#
