Compute radius of convergence Someone can help me with this radius of convergence please?
$$
\sum_{n=0}^{\infty}{\frac{3^{2}(5n^{7}+2)}{2^{n}(5n^{3}-1)}}x^{n} 
$$
I tried
$$
r = \lim_{n \to \infty } \frac{3^2(5n^7+2)(2^{n+1})(5(n+1)^{3}-1)(x^{n})}{2^{n}(5n^3-1)3^2(5(n+1)^{7}+2)(x^{n+1})}
$$
After to operate, I can't get to answer

The limit is easy to compute when you reorganize your ratio $\frac{3^2(5n^7+2)(2^{n+1})(5(n+1)^{3}-1)(x^{n})}{2^{n}(5n^3-1)3^2(5(n+1)^{7}+2)(x^{n+1})}$
as $ \frac{3^2(5n^7+2)(2^{n+1})(5(n+1)^{3}-1)(x^{n})}{2^{n}(5n^3-1)3^2(5(n+1)^{7}+2)(x^{n+1})} =\frac{5n^7+2}{5(n+1)^7+2}\cdot \frac{5(n+1)^3-1}{5n^3-1}\cdot \frac{2}{x}$.
The two first factors are rational functions, whose limit (1) is easy to compute, so it gives you $2$ for radius of convergence.
