Using the ratio test:
#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) ( (x^(n+1)/(n+1)) / (x^n/n))#
#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) n/(n+1) abs(x^(n+1)/x^n)#
#lim_(n->oo) abs(a_(n+1)/a_n) = absx lim_(n->oo) n/(n+1) =absx#
So for #absx < 1# the series is absolutely convergent and for #absx > 1# the series is not convergent.
For #absx =1# we have the two cases:
#(1) " " x=-1#:
The series becomes:
#sum_(n=0)^oo (-1)^n/n#
As:
#lim_(n->oo) abs((-1)^n/n) = lim_(n->oo) 1/n =0#
and
#abs( (-1)^(n+1)/(n+1)) = 1/(n+1) < 1/n = abs((-1)^n/n)#
the series is convergent based on Leibniz' theorem.
#(2) " " x=1#:
The series becomes:
#sum_(n=0)^oo1/n#
Consider the sequence of partial sums:
#s_n = sum_(k=1)^n 1/k#
based on Cauchy's necessary condition this sequence is convergent only if for every #epsilon >0# we can find #N# such that for every #m,n >N#:
#abs(s_n-s_m) < epsilon#
Suppose #m>n# and #p=m-n#, then:
#s_m = s_n + sum_(k=1)^p 1/(n+k)#
and:
#abs (s_n-s_m) = sum_(k=1)^p 1/(n+k)#
Now, as for #k=1,...,p# we have that: #1/(n+k) >= 1/(n+p)#
#abs (s_n-s_m) >= sum_(k=1)^p 1/(n+p)#
that is:
#abs (s_n-s_m) >= p/(n+p)#
#abs (s_n-s_m) >= 1/(1+n/p)#
So if we choose #epsilon < 1/2# and #m=2n# so that #p=n# we get:
#abs (s_n-s_m) >= 1/2 > epsilon#
which proves that Cauchy's condition is not satisfied and therefore the series is not convergent.
