Find Maclaurin's series of given function. What is its radius of convergence?
$$
g(x) = \int_0^x\frac{1}{1+t^3} dt
$$
My attempt:
$$
\int_0^x\frac{1}{1+t^3} dt = \int_0^x\frac{1}{1-(-t^3)} dt = \int_0^x\sum_{n=0}^{\infty} (-1)^n t^{3n} dt = \\
\sum_{n=0}^{\infty} (-1)^n \int_0^x t^{3n} dt = \sum_{n=0}^{\infty} (-1)^n \frac{x^{3n+1}}{3n+1}
$$
How can I find its radius of convergence? I know what to do when there is $x^n$ or $x^{2n}$ involved, but I am stuck on $x^{3n+1}$.
What also bothers me in my attempt is that I have swapped integral sign with infinite series, which I am only allowed to do when there is uniform convergence(?).
Perhaps I should just differentiate g(x):
$$
g'(x) = \frac{1}{1+x^3} \text{ which converges for } |x| < 1\\
$$
and then integrate
$$
\int\sum_{n=0}^{\infty}(-1)^nx^3n = \sum_{n=0}^{\infty} (-1)^n \frac{x^{3n+1}}{3n+1}
$$
Best Answer
The fact that the power of x is 3n+ 1 is not really a problem. Basically you find the radius of convergence using the "ratio test": the sum $\sum a_n$ converges if $\left|\frac{a_{n+1}}{a_n}\right|< 1$. Here $a_n= (-1)^n\frac{x^{3n+1}}{3n+1}$ so $a_{n+1}= (-1)^{n+1}\frac{x^{3n+ 4}}{3n+ 4}$ and $\left|\frac{a_{n+1}}{a_n}\right|= \frac{|x|^{3n+4}}{3n+4}\frac{3n+1}{|x|^{3n+1}}= \frac{3n+4}{3n+1}|x|^3$.
(I presume you know that $\frac{|x|^{3n+4}}{|x|^{3n+1}}= |x|^{3n+4-3n-1}= |x|^3$.)
That will be less than 1 if and only if $|x|^3< \frac{3n+1}{3n+4}$ so if $|x|< \sqrt[3]{\frac{3n+1}{3n+4}}$.
What is the limit of that as n goes to infinity?