Finding radius of convergence for this power summation $\sum_{n=0}^\infty \left(\int_0^n \frac{\sin^2t}{\sqrt[3]{t^7+1}} dt\right) x^n$ I have been given this tough power summation that its' general $c_n$ has an integral.
I am asked to find the radius of convergence $R$
$$\sum_{n=0}^\infty \left(\int_0^n \frac{\sin^2t}{\sqrt[3]{t^7+1}} \, dt\right) x^n $$
Do I first calculate the integral?
Any help would be appreciated!
 A: we have
$$\left|\frac{\sin^2t}{\sqrt[3]{t^7+1}}\right|\leq \frac{1}{\sqrt[3]{t^7+1}}\sim_\infty t^{-7/3}$$
and since  $\frac{7}{3}>1$ then the improper integral
$$\int_0^\infty\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt$$
is convergent and the value of the integral is $\ell\neq 0$  so if we denote by
$$a_n=\int_0^n\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt$$
then by ratio test we have
$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1$$
hence we have
$$R=1$$
A: We show that the radius of convergence is $1$. 
Let $f(x)=\frac{\sin^2 x}{\sqrt[3]{x^7+1}}$. The integral $\int_0^\infty f(x)\,dx$ converges. Suppose it has value $B$.   
Let $a_n =\int_0^n f(x)\,dx$. Then $a_n \lt B$ for all $n$.
By comparison with the series $\sum B|x|^n$, the series $\sum a_n |x|^n$ converges whenever $|x|\lt 1$. So the radius of convergence of our series is $\ge 1$.
We complete the solution by showing that the radius of convergence is $\le 1$. 
Note that $a_n \gt \int_0^1 f(x)\,dx$. Let $b$ be the value of this integral. Note that $b$ is positive. 
If $|x|\gt 1$, then $b|x|^n\to\infty$ as $n\to\infty$. So $a_n|x|^n\to\infty$ as $n\to\infty$. It follows that if $|x|\gt 1$, then $\sum a_n x^n$ diverges. 
