We start determining the radius of convergence of the series:
#sum_(n=1)^oo x^n/n^3#
using the ratio test:
#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs((x^(n+1)/((n+1)^3))/(x^n/n^3)) = lim_(n->oo) abs(x) n^3/((n+1)^3) = abs(x)#
So the ratio of convergence is #R=1#, that is the series is absolutely convergent for #abs(x) < 1#. We can easily see that the series also converges absolutely for #x =+-1# meaning that in the interval #[-1,1]# we can use the theorems for the derivation and integration term by term, and we have:
#f'(x) = d/dx sum_(n=1)^oo x^n/n^3 = sum_(n=1)^oo d/dx (x^n/n^3) = sum_(n=1)^oo (nx^(n-1))/n^3 = sum_(n=1)^oo x^(n-1)/n^2#
#int_0^x f(t)dt = int_0^x sum_(n=1)^oo t^n/n^3 dt = sum_(n=1)^oo int_0^x (t^n/n^3)dt = sum_(n=1)^oo x^(n+1)/((n+1)n^3) #
where both series have radius of convergence #R=1#. Again we can verify manually that the series also converge at the boundaries of the interval.
