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.
