<s>[INST] Find the radius of convergence and the interval of convergence of the series $\sum_{n=1}^\infty \frac{n(x-4)^n}{n^3+1}$ Series is:
$$\sum_{n=1}^\infty \frac{n(x-4)^n}{n^3+1}$$
So, I understand that I use the ratio test to find r, but I can't simplify the equation to the point where I can do this. Here's where I am so far:
$$
\frac{(n+1)(x-4)^{n+1}}{(n+1)^3+1}\bullet\frac{n^3+1}{n(x-4)^n}
$$
which gets me to
$$
\frac{(n+1)(x-4)(n^3+1)}{n[(n+1)^3+1]}
$$
Which is where I'm stuck. How can I simplify this further so I can find R? Do I just need to multiply this out and look at the coefficients of the highest powers of n?
 [/INST] $$
\lim_{n \to \infty} \left|\frac{(n+1)(x-4)(n^3+1)}{n(n^3+3n^2+3n+2)}\right| < 1 \\
\left|x-4\right|\lim_{n \to \infty} \left|\frac{(n+1)(n^3+1)}{n(n^3+3n^2+3n+2)}\right| < 1 \\
\left|x-4\right| < 1 \\
R = 1 \ \text{and} \ x \in (3, 5)
$$
</s>