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
 A: 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.
A: I recommend the root test for this one, it's a lot cleaner:
$$\lim_{n\to\infty}\left|{9^{1/n}\sqrt[n]{5n^7+2}|x|\over 2\sqrt[n]{5n^3-1}}\right|=1\cdot\left|{x\over 2}\right|\implies |x|<2\implies R=2$$
The $1$ comes from separating out the limits which go to 1 from the $|x/2|$ which doesn't.
A: If you want to use the ratio test I recommend writing the polynomial parts this way:
$${\frac{5n^7+2}{5n^3-1}}=n^4\frac{5+\frac{2}{n^7}}{5-\frac{1}{n^3}}$$
because then you have
$$\left({\frac{5n^7+2}{5n^3-1}}\right)\bigg/
  \left({\frac{5(n+1)^7+2}{5(n+1)^3-1}}\right)
  =\left(\frac{n}{n+1}\right)^4
    \left(\frac{5+\frac{2}{n^7}}{5-\frac{1}{n^3}}\right)\bigg/
    \left(\frac{5+\frac{2}{(n+1)^7}}{5-\frac{1}{(n+1)^3}}\right)$$
and all the limits you need are very easy.
