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.
