What's the radius of convergence of the next sum: $\sum_{n=0}^\infty (\int_o^n\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt)x^n$

Question

What's the radius of convergence of the next sum: $$\sum_{n=0}^\infty \left(\int_0^n\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt\right)x^n$$

I know that $$\int_0^\infty\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt$$ does converge:

Let $$a(n)=\frac{\sin^2t}{\sqrt[3]{t^7+1}}\le \frac 1{\sqrt[3]{t^7}}=b(n),$$ $b(n)$ converges using integral test.

But now I'm stuck, How can I find that radius of convergence?

Answer 1

Let $a_n=\int_0^n {\sin^2 t\over \root 3\of{t^7+1}}\,dt$. You've noted that $(a_n)$ converges. Now note it must converge to some number $L\ne0$. Then $(a_{n+1})$ does the same.

So, what's $\lim\limits_{n\rightarrow\infty}{a_n\over a_{n+1}}$?

Answer 2

If $(a_n,n\geqslant 1)$ is a bounded sequence, then the radius of convergence of $\sum_na_nx^n$ is at least $1$. Here, it cannot be strictly more than $1$ since $a_n\geqslant\int_0^1\frac{\sin^2t}{\sqrt[3]{t^2+1}}dt$ for all $n$.